OP Malhotra Class 12 Maths Solutions Chapter 22 Vectors Exercise 22 (A)

Question 1. If \( \vec{a} = \hat{i} + 2 \hat{j} – \hat{k} \), \( \vec{b} = 2 \hat{i} + \hat{j} + \hat{k} \) and \( \vec{c} = 5 \hat{i} – 4 \hat{j} + 3 \hat{k} \), then find the value of
(i) \( \vec{a} \cdot \vec{b} \)
(ii) \( \vec{b} \cdot \vec{c} \)
(iii) \( \vec{c} \cdot \vec{a} \)
(iv) \( |\vec{a}| \)
(v) \( |\vec{b}| \)
(vi) \( |\vec{c}| \)
(vii) \( \vec{a} (\vec{b} + \vec{c}), \vec{a} \vec{b} + \vec{a} \vec{c} \) and show that they are equal.
(viii) \( (2 \vec{a} – \vec{b} + 2 \vec{c}) \cdot \vec{c} \)
(ix) \( (\vec{a} - 2 \vec{b}) \cdot (\vec{b} – 2 \vec{c}) \)
(x) angle between \( \vec{a} \) and \( \vec{b} \), \( \vec{b} \) and \( \vec{c} \) and \( \vec{c} \) and \( \vec{a} \).
Answer:
Given vectors:
\( \vec{a} = \hat{i} + 2 \hat{j} – \hat{k} \)
\( \vec{b} = 2 \hat{i} + \hat{j} + \hat{k} \)
\( \vec{c} = 5 \hat{i} – 4 \hat{j} + 3 \hat{k} \)

(i) To find \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (\hat{i} + 2 \hat{j} – \hat{k}) \cdot (2 \hat{i} + \hat{j} + \hat{k}) \)
This is found by multiplying the corresponding components and adding them.
\( = 1(2) + 2(1) + (-1)(1) \)
\( = 2 + 2 - 1 \)
\( = 3 \)

(ii) To find \( \vec{b} \cdot \vec{c} \):
\( \vec{b} \cdot \vec{c} = (2 \hat{i} + \hat{j} + \hat{k}) \cdot (5 \hat{i} – 4 \hat{j} + 3 \hat{k}) \)
\( = 2(5) + 1(-4) + 1(3) \)
\( = 10 - 4 + 3 \)
\( = 9 \)

(iii) To find \( \vec{c} \cdot \vec{a} \):
\( \vec{c} \cdot \vec{a} = (5 \hat{i} – 4 \hat{j} + 3 \hat{k}) \cdot (\hat{i} + 2 \hat{j} – \hat{k}) \)
\( = 5(1) + (-4)(2) + 3(-1) \)
\( = 5 - 8 - 3 \)
\( = -6 \)

(iv) To find \( |\vec{a}| \):
The magnitude of a vector is the square root of the sum of the squares of its components.
\( |\vec{a}| = \sqrt{1^2 + 2^2 + (-1)^2} \)
\( = \sqrt{1 + 4 + 1} \)
\( = \sqrt{6} \)

(v) To find \( |\vec{b}| \):
\( |\vec{b}| = \sqrt{2^2 + 1^2 + 1^2} \)
\( = \sqrt{4 + 1 + 1} \)
\( = \sqrt{6} \)

(vi) To find \( |\vec{c}| \):
\( |\vec{c}| = \sqrt{5^2 + (-4)^2 + 3^2} \)
\( = \sqrt{25 + 16 + 9} \)
\( = \sqrt{50} \)
\( = 5 \sqrt{2} \)

(vii) To verify \( \vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \):
First, calculate \( \vec{b} + \vec{c} \):
\( \vec{b} + \vec{c} = (2 \hat{i} + \hat{j} + \hat{k}) + (5 \hat{i} – 4 \hat{j} + 3 \hat{k}) \)
\( = (2+5)\hat{i} + (1-4)\hat{j} + (1+3)\hat{k} \)
\( = 7 \hat{i} – 3 \hat{j} + 4 \hat{k} \)
Next, calculate \( \vec{a} \cdot (\vec{b} + \vec{c}) \):
\( \vec{a} \cdot (\vec{b} + \vec{c}) = (\hat{i} + 2 \hat{j} – \hat{k}) \cdot (7 \hat{i} – 3 \hat{j} + 4 \hat{k}) \)
\( = 1(7) + 2(-3) + (-1)(4) \)
\( = 7 - 6 - 4 \)
\( = -3 \)
From parts (i) and (iii), we have \( \vec{a} \cdot \vec{b} = 3 \) and \( \vec{a} \cdot \vec{c} = -6 \).
So, \( \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 3 + (-6) = -3 \)
Since both results are -3, it is shown that \( \vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \). This property is called the distributive property of the dot product over vector addition.

(viii) To find \( (2 \vec{a} – \vec{b} + 2 \vec{c}) \cdot \vec{c} \):
First, calculate the vector \( 2 \vec{a} – \vec{b} + 2 \vec{c} \):
\( 2 \vec{a} = 2(\hat{i} + 2 \hat{j} – \hat{k}) = 2 \hat{i} + 4 \hat{j} – 2 \hat{k} \)
\( 2 \vec{c} = 2(5 \hat{i} – 4 \hat{j} + 3 \hat{k}) = 10 \hat{i} – 8 \hat{j} + 6 \hat{k} \)
Now, \( 2 \vec{a} – \vec{b} + 2 \vec{c} \)
\( = (2 \hat{i} + 4 \hat{j} – 2 \hat{k}) - (2 \hat{i} + \hat{j} + \hat{k}) + (10 \hat{i} – 8 \hat{j} + 6 \hat{k}) \)
\( = (2 - 2 + 10)\hat{i} + (4 - 1 - 8)\hat{j} + (-2 - 1 + 6)\hat{k} \)
\( = 10 \hat{i} – 5 \hat{j} + 3 \hat{k} \)
Then, find the dot product with \( \vec{c} \):
\( (10 \hat{i} – 5 \hat{j} + 3 \hat{k}) \cdot (5 \hat{i} – 4 \hat{j} + 3 \hat{k}) \)
\( = 10(5) + (-5)(-4) + 3(3) \)
\( = 50 + 20 + 9 \)
\( = 79 \)

(ix) To find \( (\vec{a} - 2 \vec{b}) \cdot (\vec{b} – 2 \vec{c}) \):
First, calculate \( \vec{a} - 2 \vec{b} \):
\( 2 \vec{b} = 2(2 \hat{i} + \hat{j} + \hat{k}) = 4 \hat{i} + 2 \hat{j} + 2 \hat{k} \)
\( \vec{a} - 2 \vec{b} = (\hat{i} + 2 \hat{j} – \hat{k}) - (4 \hat{i} + 2 \hat{j} + 2 \hat{k}) \)
\( = (1 - 4)\hat{i} + (2 - 2)\hat{j} + (-1 - 2)\hat{k} \)
\( = -3 \hat{i} + 0 \hat{j} – 3 \hat{k} \)
Next, calculate \( \vec{b} – 2 \vec{c} \):
\( 2 \vec{c} = 2(5 \hat{i} – 4 \hat{j} + 3 \hat{k}) = 10 \hat{i} – 8 \hat{j} + 6 \hat{k} \)
\( \vec{b} – 2 \vec{c} = (2 \hat{i} + \hat{j} + \hat{k}) - (10 \hat{i} – 8 \hat{j} + 6 \hat{k}) \)
\( = (2 - 10)\hat{i} + (1 - (-8))\hat{j} + (1 - 6)\hat{k} \)
\( = -8 \hat{i} + 9 \hat{j} – 5 \hat{k} \)
Then, find the dot product of these two vectors:
\( (\vec{a} - 2 \vec{b}) \cdot (\vec{b} – 2 \vec{c}) = (-3 \hat{i} – 3 \hat{k}) \cdot (-8 \hat{i} + 9 \hat{j} – 5 \hat{k}) \)
\( = (-3)(-8) + (0)(9) + (-3)(-5) \)
\( = 24 + 0 + 15 \)
\( = 39 \)

(x) To find the angles between the vector pairs:
The angle \( \theta \) between two vectors \( \vec{A} \) and \( \vec{B} \) is given by \( \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} \).
Let \( \theta_1 \) be the angle between \( \vec{a} \) and \( \vec{b} \).
Using values from (i), (iv), and (v):
\( \vec{a} \cdot \vec{b} = 3 \)
\( |\vec{a}| = \sqrt{6} \)
\( |\vec{b}| = \sqrt{6} \)
\( \cos \theta_1 = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2} \)
\( \implies \theta_1 = \cos^{-1} \left( \frac{1}{2} \right) \)
\( \implies \theta_1 = \frac{\pi}{3} \) or 60 degrees. This is a common angle in geometry.

Let \( \theta_2 \) be the angle between \( \vec{b} \) and \( \vec{c} \).
Using values from (ii), (v), and (vi):
\( \vec{b} \cdot \vec{c} = 9 \)
\( |\vec{b}| = \sqrt{6} \)
\( |\vec{c}| = 5 \sqrt{2} \)
\( \cos \theta_2 = \frac{9}{\sqrt{6} \cdot 5 \sqrt{2}} = \frac{9}{5 \sqrt{12}} = \frac{9}{5 \cdot 2 \sqrt{3}} = \frac{9}{10 \sqrt{3}} \)
\( \implies \theta_2 = \cos^{-1} \left( \frac{9}{10 \sqrt{3}} \right) \)

Let \( \theta_3 \) be the angle between \( \vec{c} \) and \( \vec{a} \).
Using values from (iii), (iv), and (vi):
\( \vec{c} \cdot \vec{a} = -6 \)
\( |\vec{c}| = 5 \sqrt{2} \)
\( |\vec{a}| = \sqrt{6} \)
\( \cos \theta_3 = \frac{-6}{5 \sqrt{2} \cdot \sqrt{6}} = \frac{-6}{5 \sqrt{12}} = \frac{-6}{5 \cdot 2 \sqrt{3}} = \frac{-6}{10 \sqrt{3}} = \frac{-3}{5 \sqrt{3}} \)
To rationalize the denominator, multiply top and bottom by \( \sqrt{3} \):
\( = \frac{-3 \sqrt{3}}{5 \cdot 3} = \frac{-\sqrt{3}}{5} \)
\( \implies \theta_3 = \cos^{-1} \left( \frac{-\sqrt{3}}{5} \right) \)
In simple words: For finding the dot product, multiply the matching parts of the two vectors (x with x, y with y, z with z) and add them up. For magnitude, square each part, add them, and take the square root. To find the angle, use the formula that connects the dot product and magnitudes. Remember that the cosine of an angle gives a good idea of its size.

🎯 Exam Tip: Pay close attention to the signs when performing dot products and vector subtractions. A single sign error can lead to an incorrect final answer. Ensure all steps are shown clearly for partial credit.

Question 2. If \( \vec{a} = \hat{i} + 2 \hat{j} + \hat{k} \), \( \vec{b} = 3 \hat{i} – \hat{j} + 4 \hat{k} \) and \( \vec{c} = \hat{i} – \hat{j} + \hat{k} \) show that \( \vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \).
Answer:
Given vectors:
\( \vec{a} = \hat{i} + 2 \hat{j} + \hat{k} \)
\( \vec{b} = 3 \hat{i} – \hat{j} + 4 \hat{k} \)
\( \vec{c} = \hat{i} – \hat{j} + \hat{k} \)

First, let's find the Left Hand Side (L.H.S.): \( \vec{a} \cdot (\vec{b} + \vec{c}) \).
Calculate \( \vec{b} + \vec{c} \):
\( \vec{b} + \vec{c} = (3 \hat{i} – \hat{j} + 4 \hat{k}) + (\hat{i} – \hat{j} + \hat{k}) \)
\( = (3+1)\hat{i} + (-1-1)\hat{j} + (4+1)\hat{k} \)
\( = 4 \hat{i} – 2 \hat{j} + 5 \hat{k} \)
Now, calculate \( \vec{a} \cdot (\vec{b} + \vec{c}) \):
L.H.S. \( = (\hat{i} + 2 \hat{j} + \hat{k}) \cdot (4 \hat{i} – 2 \hat{j} + 5 \hat{k}) \)
\( = 1(4) + 2(-2) + 1(5) \)
\( = 4 - 4 + 5 \)
\( = 5 \)

Next, let's find the Right Hand Side (R.H.S.): \( \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \).
Calculate \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (\hat{i} + 2 \hat{j} + \hat{k}) \cdot (3 \hat{i} – \hat{j} + 4 \hat{k}) \)
\( = 1(3) + 2(-1) + 1(4) \)
\( = 3 - 2 + 4 \)
\( = 5 \)
Calculate \( \vec{a} \cdot \vec{c} \):
\( \vec{a} \cdot \vec{c} = (\hat{i} + 2 \hat{j} + \hat{k}) \cdot (\hat{i} – \hat{j} + \hat{k}) \)
\( = 1(1) + 2(-1) + 1(1) \)
\( = 1 - 2 + 1 \)
\( = 0 \)
Now, add the two dot products:
R.H.S. \( = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = 5 + 0 \)
\( = 5 \)

Since L.H.S. = 5 and R.H.S. = 5, we have shown that \( \vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} \). This is a fundamental property in vector algebra, proving the distributive nature of the dot product.
In simple words: First, add vector b and c together, then find the dot product of vector a with this new sum. This gives you one number. Then, find the dot product of a with b, and the dot product of a with c, and add those two numbers. Both final numbers should be the same.

🎯 Exam Tip: Always calculate both sides of the equation separately and clearly show each step. This method helps to organize your work and spot any potential calculation errors quickly.

Question 3. Compute the angle between \( \vec{a} \) and \( \vec{b} \), where
(i) \( \vec{a} = 3 \hat{i} + \hat{j} + 2 \hat{k} \) and \( \vec{b} = 2 \hat{i} – 2 \hat{j} + 4 \hat{k} \)
(ii) \( \vec{a} = 2 \hat{i} + 3 \hat{j} \) and \( \vec{b} = 2 \hat{i} + 3 \hat{k} \)
(iii) \( \vec{a} = 3 \hat{i} – 2 \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} – 2 \hat{j} – 3 \hat{k} \).
Answer:
The angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is found using the formula: \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \).

(i) Given \( \vec{a} = 3 \hat{i} + \hat{j} + 2 \hat{k} \) and \( \vec{b} = 2 \hat{i} – 2 \hat{j} + 4 \hat{k} \).
First, find \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (3)(2) + (1)(-2) + (2)(4) \)
\( = 6 - 2 + 8 \)
\( = 12 \)
Next, find the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \):
\( |\vec{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \)
\( |\vec{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \)
Now, use the cosine formula:
\( \cos \theta = \frac{12}{\sqrt{14} \sqrt{24}} = \frac{12}{\sqrt{14 \times 24}} = \frac{12}{\sqrt{336}} \)
Simplify \( \sqrt{336} \): \( \sqrt{336} = \sqrt{16 \times 21} = 4\sqrt{21} \)
\( \cos \theta = \frac{12}{4\sqrt{21}} = \frac{3}{\sqrt{21}} \)
To rationalize, multiply by \( \frac{\sqrt{21}}{\sqrt{21}} \): \( = \frac{3\sqrt{21}}{21} = \frac{\sqrt{21}}{7} \)
\( \implies \theta = \cos^{-1} \left( \frac{\sqrt{21}}{7} \right) \)

(ii) Given \( \vec{a} = 2 \hat{i} + 3 \hat{j} \) (which means \( 2 \hat{i} + 3 \hat{j} + 0 \hat{k} \)) and \( \vec{b} = 2 \hat{i} + 3 \hat{k} \) (which means \( 2 \hat{i} + 0 \hat{j} + 3 \hat{k} \)).
First, find \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (2)(2) + (3)(0) + (0)(3) \)
\( = 4 + 0 + 0 \)
\( = 4 \)
Next, find the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \):
\( |\vec{a}| = \sqrt{2^2 + 3^2 + 0^2} = \sqrt{4 + 9} = \sqrt{13} \)
\( |\vec{b}| = \sqrt{2^2 + 0^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \)
Now, use the cosine formula:
\( \cos \theta = \frac{4}{\sqrt{13} \cdot \sqrt{13}} = \frac{4}{13} \)
\( \implies \theta = \cos^{-1} \left( \frac{4}{13} \right) \)

(iii) Given \( \vec{a} = 3 \hat{i} – 2 \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} – 2 \hat{j} – 3 \hat{k} \).
First, find \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (3)(1) + (-2)(-2) + (1)(-3) \)
\( = 3 + 4 - 3 \)
\( = 4 \)
Next, find the magnitudes \( |\vec{a}| \) and \( |\vec{b}| \):
\( |\vec{a}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \)
\( |\vec{b}| = \sqrt{1^2 + (-2)^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \)
Now, use the cosine formula:
\( \cos \theta = \frac{4}{\sqrt{14} \cdot \sqrt{14}} = \frac{4}{14} = \frac{2}{7} \)
\( \implies \theta = \cos^{-1} \left( \frac{2}{7} \right) \)
In simple words: To find the angle between two vectors, first multiply their matching components and add them up (dot product). Then, find the length of each vector (magnitude). Finally, divide the dot product by the product of their lengths, and use the inverse cosine function to get the angle.

🎯 Exam Tip: Remember that \( \cos^{-1} \) often gives an angle in radians, but it's also acceptable to express it in degrees if specified. Make sure to simplify any square roots and fractions for the clearest answer.

Question 4. Prove that the following pair of vectors are perpendicular to each other :
(i) \( \vec{a} = \hat{i} + 2 \hat{j} – 4 \hat{k} \) and \( \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} \)
(ii) \( \overrightarrow{\mathbf{A}} = a \hat{i} + b \hat{j} + c \hat{k} \) and \( \overrightarrow{\mathbf{B}} = (b – c) \hat{i} + (c – a) \hat{j} + (a – b) \hat{k} \)
Answer:
Two vectors are perpendicular (orthogonal) if their dot product is zero.

(i) Given \( \vec{a} = \hat{i} + 2 \hat{j} – 4 \hat{k} \) and \( \vec{b} = \hat{i} + 2 \hat{j} + \hat{k} \).
Calculate the dot product \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (1)(1) + (2)(2) + (-4)(1) \)
\( = 1 + 4 - 4 \)
\( = 1 \)
Since \( \vec{a} \cdot \vec{b} = 1 \), which is not zero, these vectors are **not** perpendicular. There might be a typo in the question or the given answer implies they should be perpendicular. However, based on the calculation, they are not. If the intent was for them to be perpendicular, the problem statement would need adjustment.

(ii) Given \( \overrightarrow{\mathbf{A}} = a \hat{i} + b \hat{j} + c \hat{k} \) and \( \overrightarrow{\mathbf{B}} = (b – c) \hat{i} + (c – a) \hat{j} + (a – b) \hat{k} \).
Calculate the dot product \( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} \):
\( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = a(b – c) + b(c – a) + c(a – b) \)
\( = ab - ac + bc - ba + ca - cb \)
Rearrange the terms:
\( = (ab - ba) + (bc - cb) + (ca - ac) \)
\( = 0 + 0 + 0 \)
\( = 0 \)
Since \( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = 0 \), the vectors \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{B}} \) are perpendicular to each other. This shows a clever arrangement of components that naturally results in a zero dot product.
In simple words: To prove that two vectors are perpendicular, you just need to multiply their corresponding parts (like x-parts together, y-parts together, etc.) and add up the results. If this total sum is zero, then the vectors are perpendicular, meaning they form a right angle with each other.

🎯 Exam Tip: The condition for perpendicularity is a dot product of zero. For algebraic vector components, be very careful with signs and expand all terms correctly to ensure they cancel out as expected.

Question 5.
(i) If \( |\vec{a}| = 2 \), \( |\vec{b}| = 3 \) and \( \vec{a} \cdot \vec{b} = 3 \), find the angle between \( \vec{a} \) and \( \vec{b} \).
(ii) If \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \) and \( \vec{a} \cdot \vec{b} = 0 \), find the angle between the two vectors.
Answer:
The angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is found using the formula: \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \).

(i) Given \( |\vec{a}| = 2 \), \( |\vec{b}| = 3 \) and \( \vec{a} \cdot \vec{b} = 3 \).
Substitute these values into the formula:
\( \cos \theta = \frac{3}{2 \times 3} \)
\( \cos \theta = \frac{3}{6} \)
\( \cos \theta = \frac{1}{2} \)
To find \( \theta \), take the inverse cosine of \( \frac{1}{2} \):
\( \implies \theta = \frac{\pi}{3} \) radians or \( 60^\circ \). This is a common and easily recognizable angle.

(ii) Given \( |\vec{a}| = 1 \), \( |\vec{b}| = 2 \) and \( \vec{a} \cdot \vec{b} = 0 \).
Substitute these values into the formula:
\( \cos \theta = \frac{0}{1 \times 2} \)
\( \cos \theta = \frac{0}{2} \)
\( \cos \theta = 0 \)
To find \( \theta \), take the inverse cosine of \( 0 \):
\( \implies \theta = \frac{\pi}{2} \) radians or \( 90^\circ \). A zero dot product always means the vectors are perpendicular.
In simple words: To find the angle between vectors, you need their dot product and their lengths. Put these numbers into the angle formula and then use a calculator or your knowledge of angles to find the degree or radian measure. If the dot product is zero, the angle is always 90 degrees.

🎯 Exam Tip: Remember the special case: if the dot product of two non-zero vectors is zero, the vectors are perpendicular, and the angle between them is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.

Question 6. If A is (1, 3, 2), B is (-1, 1, 1) and C is (-3, 2, 3), prove that \( \overrightarrow{AB} \cdot \overrightarrow{BC} = 0 \). What can be said about \( \triangle ABC \)?
Answer:
Given the position vectors (P.V.) of points A, B, and C:
P.V. of A \( = \hat{i} + 3 \hat{j} + 2 \hat{k} \)
P.V. of B \( = -\hat{i} + \hat{j} + \hat{k} \)
P.V. of C \( = -3 \hat{i} + 2 \hat{j} + 3 \hat{k} \)

First, find the vector \( \overrightarrow{AB} \):
\( \overrightarrow{AB} = \text{P.V. of B} - \text{P.V. of A} \)
\( = (-\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 3 \hat{j} + 2 \hat{k}) \)
\( = (-1-1)\hat{i} + (1-3)\hat{j} + (1-2)\hat{k} \)
\( = -2 \hat{i} – 2 \hat{j} – \hat{k} \)

Next, find the vector \( \overrightarrow{BC} \):
\( \overrightarrow{BC} = \text{P.V. of C} - \text{P.V. of B} \)
\( = (-3 \hat{i} + 2 \hat{j} + 3 \hat{k}) - (-\hat{i} + \hat{j} + \hat{k}) \)
\( = (-3-(-1))\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} \)
\( = -2 \hat{i} + \hat{j} + 2 \hat{k} \)

Now, calculate the dot product \( \overrightarrow{AB} \cdot \overrightarrow{BC} \):
\( \overrightarrow{AB} \cdot \overrightarrow{BC} = (-2 \hat{i} – 2 \hat{j} – \hat{k}) \cdot (-2 \hat{i} + \hat{j} + 2 \hat{k}) \)
\( = (-2)(-2) + (-2)(1) + (-1)(2) \)
\( = 4 - 2 - 2 \)
\( = 0 \)

Since \( \overrightarrow{AB} \cdot \overrightarrow{BC} = 0 \), the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) are perpendicular to each other. This means that the angle between these two vectors is \( 90^\circ \). Therefore, \( \triangle ABC \) is a right-angled triangle with the right angle at vertex B. Vector dot products are very useful for identifying right angles in geometric shapes.
In simple words: First, find the vectors from point A to B and from point B to C. Then, multiply their matching parts and add them up. If the total sum is zero, it means these two vectors meet at a 90-degree angle. This tells us that the triangle ABC has a right angle at point B.

🎯 Exam Tip: When dealing with points and vectors, always start by defining the position vectors of the points. Then, carefully calculate the vectors representing the sides of the triangle before computing their dot product.

Question 7.
(i) Given \( \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = 3 \hat{i} – \hat{j} + 2 \hat{k} \), show that \( (\vec{a} + \vec{b}) \) is perpendicular to \( (\vec{a} – \vec{b}) \).
(ii) Find the value of x so that the pairs of vectors \( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \) and \( 4 \hat{i} – 2 \hat{j} + x \hat{k} \) may be perpendicular to each other.
(iii) If \( \vec{a} = (5 \hat{i} – \hat{j} + 7 \hat{k}) \) and \( \vec{b} = (\hat{i} – \hat{j} – \lambda \hat{k}) \), find the value of \( \lambda \) for which \( (\vec{a} + \vec{b}) \) and \( (\vec{a} – \vec{b}) \) are orthogonal.
(iv) Find the angle between the vectors \( (\vec{a} + \vec{b}) \) and \( (\vec{a} – \vec{b}) \) for \( \vec{a} = 2 \hat{i} – \hat{j} + 3 \hat{k} \) and \( \vec{b} = 3 \hat{i} + \hat{j} – 2 \hat{k} \).
Answer:
Two vectors are perpendicular (orthogonal) if their dot product is zero.

(i) Given \( \vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \) and \( \vec{b} = 3 \hat{i} – \hat{j} + 2 \hat{k} \).
First, find \( \vec{a} + \vec{b} \):
\( \vec{a} + \vec{b} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + (3 \hat{i} – \hat{j} + 2 \hat{k}) \)
\( = (1+3)\hat{i} + (2-1)\hat{j} + (3+2)\hat{k} \)
\( = 4 \hat{i} + \hat{j} + 5 \hat{k} \)
Next, find \( \vec{a} – \vec{b} \):
\( \vec{a} – \vec{b} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} – \hat{j} + 2 \hat{k}) \)
\( = (1-3)\hat{i} + (2-(-1))\hat{j} + (3-2)\hat{k} \)
\( = -2 \hat{i} + 3 \hat{j} + \hat{k} \)
Now, calculate the dot product \( (\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) \):
\( (\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = (4 \hat{i} + \hat{j} + 5 \hat{k}) \cdot (-2 \hat{i} + 3 \hat{j} + \hat{k}) \)
\( = (4)(-2) + (1)(3) + (5)(1) \)
\( = -8 + 3 + 5 \)
\( = 0 \)
Since the dot product is 0, \( (\vec{a} + \vec{b}) \) is perpendicular to \( (\vec{a} – \vec{b}) \). This is a general property: for any vectors \( \vec{a} \) and \( \vec{b} \), if \( |\vec{a}| = |\vec{b}| \), then \( \vec{a} + \vec{b} \) and \( \vec{a} – \vec{b} \) are perpendicular. Here, they are not equal, but the specific components lead to a zero dot product anyway.

(ii) Let the two vectors be \( \vec{u} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \) and \( \vec{v} = 4 \hat{i} – 2 \hat{j} + x \hat{k} \).
For them to be perpendicular, their dot product must be zero: \( \vec{u} \cdot \vec{v} = 0 \).
\( (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (4 \hat{i} – 2 \hat{j} + x \hat{k}) = 0 \)
\( (2)(4) + (3)(-2) + (4)(x) = 0 \)
\( 8 - 6 + 4x = 0 \)
\( 2 + 4x = 0 \)
\( 4x = -2 \)
\( x = -\frac{2}{4} \)
\( x = -\frac{1}{2} \)
The value of x for which the vectors are perpendicular is \( -\frac{1}{2} \).

(iii) Given \( \vec{a} = 5 \hat{i} – \hat{j} + 7 \hat{k} \) and \( \vec{b} = \hat{i} – \hat{j} – \lambda \hat{k} \).
For \( (\vec{a} + \vec{b}) \) and \( (\vec{a} – \vec{b}) \) to be orthogonal, their dot product must be zero.
First, find \( \vec{a} + \vec{b} \):
\( \vec{a} + \vec{b} = (5 \hat{i} – \hat{j} + 7 \hat{k}) + (\hat{i} – \hat{j} – \lambda \hat{k}) \)
\( = (5+1)\hat{i} + (-1-1)\hat{j} + (7-\lambda)\hat{k} \)
\( = 6 \hat{i} – 2 \hat{j} + (7 – \lambda) \hat{k} \)
Next, find \( \vec{a} – \vec{b} \):
\( \vec{a} – \vec{b} = (5 \hat{i} – \hat{j} + 7 \hat{k}) - (\hat{i} – \hat{j} – \lambda \hat{k}) \)
\( = (5-1)\hat{i} + (-1-(-1))\hat{j} + (7-(-\lambda))\hat{k} \)
\( = 4 \hat{i} + 0 \hat{j} + (7 + \lambda) \hat{k} \)
Now, set their dot product to zero:
\( (\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = 0 \)
\( (6 \hat{i} – 2 \hat{j} + (7 – \lambda) \hat{k}) \cdot (4 \hat{i} + 0 \hat{j} + (7 + \lambda) \hat{k}) = 0 \)
\( (6)(4) + (-2)(0) + (7 – \lambda)(7 + \lambda) = 0 \)
\( 24 + 0 + (49 – \lambda^2) = 0 \)
\( 24 + 49 – \lambda^2 = 0 \)
\( 73 – \lambda^2 = 0 \)
\( \lambda^2 = 73 \)
\( \lambda = \pm \sqrt{73} \)
So, there are two possible values for \( \lambda \).

(iv) Given \( \vec{a} = 2 \hat{i} – \hat{j} + 3 \hat{k} \) and \( \vec{b} = 3 \hat{i} + \hat{j} – 2 \hat{k} \).
We need to find the angle between \( (\vec{a} + \vec{b}) \) and \( (\vec{a} – \vec{b}) \).
First, find \( \vec{a} + \vec{b} \):
\( \vec{a} + \vec{b} = (2 \hat{i} – \hat{j} + 3 \hat{k}) + (3 \hat{i} + \hat{j} – 2 \hat{k}) \)
\( = (2+3)\hat{i} + (-1+1)\hat{j} + (3-2)\hat{k} \)
\( = 5 \hat{i} + 0 \hat{j} + \hat{k} \)
Next, find \( \vec{a} – \vec{b} \):
\( \vec{a} – \vec{b} = (2 \hat{i} – \hat{j} + 3 \hat{k}) - (3 \hat{i} + \hat{j} – 2 \hat{k}) \)
\( = (2-3)\hat{i} + (-1-1)\hat{j} + (3-(-2))\hat{k} \)
\( = -\hat{i} – 2 \hat{j} + 5 \hat{k} \)
Now, find the dot product of \( (\vec{a} + \vec{b}) \) and \( (\vec{a} – \vec{b}) \):
\( (\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = (5 \hat{i} + \hat{k}) \cdot (-\hat{i} – 2 \hat{j} + 5 \hat{k}) \)
\( = (5)(-1) + (0)(-2) + (1)(5) \)
\( = -5 + 0 + 5 \)
\( = 0 \)
Since the dot product is 0, the angle between \( (\vec{a} + \vec{b}) \) and \( (\vec{a} – \vec{b}) \) is \( \frac{\pi}{2} \) radians or \( 90^\circ \). This often happens when vectors have equal magnitudes, but the dot product confirms it.
In simple words: To check if vectors are perpendicular, or to find a missing value that makes them perpendicular, just find their dot product and set it to zero. If you need to find the angle, calculate the dot product and the magnitudes, then use the cosine formula. Perpendicular vectors always make a 90-degree angle.

🎯 Exam Tip: Remember that \( (\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2 \). So, if this dot product is zero, it implies \( |\vec{a}|^2 = |\vec{b}|^2 \), meaning \( |\vec{a}| = |\vec{b}| \). This is a useful shortcut to remember for such problems.

Question 8. Find the angles of the triangle whose vertices are A(0, -1, -2), B(3, 1, 4) and C(5, 7, 1).
Answer:
Given the vertices of the triangle A(0, -1, -2), B(3, 1, 4), and C(5, 7, 1).
First, write the position vectors (P.V.) for each point:
P.V. of A \( = 0 \hat{i} – \hat{j} – 2 \hat{k} \)
P.V. of B \( = 3 \hat{i} + \hat{j} + 4 \hat{k} \)
P.V. of C \( = 5 \hat{i} + 7 \hat{j} + \hat{k} \)

Next, find the vectors representing the sides of the triangle:
\( \overrightarrow{AB} = \text{P.V. of B} - \text{P.V. of A} \)
\( = (3 \hat{i} + \hat{j} + 4 \hat{k}) - (0 \hat{i} – \hat{j} – 2 \hat{k}) \)
\( = (3-0)\hat{i} + (1-(-1))\hat{j} + (4-(-2))\hat{k} \)
\( = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \)
\( \overrightarrow{BC} = \text{P.V. of C} - \text{P.V. of B} \)
\( = (5 \hat{i} + 7 \hat{j} + \hat{k}) - (3 \hat{i} + \hat{j} + 4 \hat{k}) \)
\( = (5-3)\hat{i} + (7-1)\hat{j} + (1-4)\hat{k} \)
\( = 2 \hat{i} + 6 \hat{j} – 3 \hat{k} \)
\( \overrightarrow{CA} = \text{P.V. of A} - \text{P.V. of C} \)
\( = (0 \hat{i} – \hat{j} – 2 \hat{k}) - (5 \hat{i} + 7 \hat{j} + \hat{k}) \)
\( = (0-5)\hat{i} + (-1-7)\hat{j} + (-2-1)\hat{k} \)
\( = -5 \hat{i} – 8 \hat{j} – 3 \hat{k} \)

Now, find the magnitudes of these vectors:
\( |\overrightarrow{AB}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \)
\( |\overrightarrow{BC}| = \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \)
\( |\overrightarrow{CA}| = \sqrt{(-5)^2 + (-8)^2 + (-3)^2} = \sqrt{25 + 64 + 9} = \sqrt{98} \)
\( = \sqrt{49 \times 2} = 7\sqrt{2} \)
Since \( |\overrightarrow{AB}| = |\overrightarrow{BC}| \), this is an isosceles triangle.

To find the angles of the triangle (A, B, C), we use the dot product formula \( \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \). For an interior angle, the vectors must point either both away from or both towards the vertex.

Angle at A: Consider vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \). Note that \( \overrightarrow{AC} = - \overrightarrow{CA} \).
\( \overrightarrow{AC} = - (-5 \hat{i} – 8 \hat{j} – 3 \hat{k}) = 5 \hat{i} + 8 \hat{j} + 3 \hat{k} \)
\( \cos A = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} \)
\( \overrightarrow{AB} \cdot \overrightarrow{AC} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) \cdot (5 \hat{i} + 8 \hat{j} + 3 \hat{k}) \)
\( = 3(5) + 2(8) + 6(3) \)
\( = 15 + 16 + 18 \)
\( = 49 \)
\( |\overrightarrow{AC}| = |\overrightarrow{CA}| = 7\sqrt{2} \)
\( \cos A = \frac{49}{7 \cdot 7\sqrt{2}} = \frac{49}{49\sqrt{2}} = \frac{1}{\sqrt{2}} \)
\( \implies A = \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} \) or \( 45^\circ \).

Angle at B: Consider vectors \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \). Note that \( \overrightarrow{BA} = - \overrightarrow{AB} \).
\( \overrightarrow{BA} = - (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = -3 \hat{i} – 2 \hat{j} – 6 \hat{k} \)
\( \cos B = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} \)
\( \overrightarrow{BA} \cdot \overrightarrow{BC} = (-3 \hat{i} – 2 \hat{j} – 6 \hat{k}) \cdot (2 \hat{i} + 6 \hat{j} – 3 \hat{k}) \)
\( = (-3)(2) + (-2)(6) + (-6)(-3) \)
\( = -6 - 12 + 18 \)
\( = 0 \)
Since the dot product is 0, \( \cos B = 0 \).
\( \implies B = \cos^{-1}(0) = \frac{\pi}{2} \) or \( 90^\circ \). This means it is a right-angled triangle.

Angle at C: Consider vectors \( \overrightarrow{CA} \) and \( \overrightarrow{CB} \). Note that \( \overrightarrow{CB} = - \overrightarrow{BC} \).
\( \overrightarrow{CB} = - (2 \hat{i} + 6 \hat{j} – 3 \hat{k}) = -2 \hat{i} – 6 \hat{j} + 3 \hat{k} \)
\( \cos C = \frac{\overrightarrow{CA} \cdot \overrightarrow{CB}}{|\overrightarrow{CA}||\overrightarrow{CB}|} \)
\( \overrightarrow{CA} \cdot \overrightarrow{CB} = (-5 \hat{i} – 8 \hat{j} – 3 \hat{k}) \cdot (-2 \hat{i} – 6 \hat{j} + 3 \hat{k}) \)
\( = (-5)(-2) + (-8)(-6) + (-3)(3) \)
\( = 10 + 48 - 9 \)
\( = 49 \)
\( |\overrightarrow{CB}| = |\overrightarrow{BC}| = 7 \)
\( \cos C = \frac{49}{7\sqrt{2} \cdot 7} = \frac{49}{49\sqrt{2}} = \frac{1}{\sqrt{2}} \)
\( \implies C = \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} \) or \( 45^\circ \).

The angles of the triangle are \( A = 45^\circ \), \( B = 90^\circ \), and \( C = 45^\circ \). The sum of the angles is \( 45^\circ + 90^\circ + 45^\circ = 180^\circ \), which confirms the calculations. This is a right-angled isosceles triangle.
In simple words: To find the angles of a triangle when you have the corner points, first find the vectors for each side. Then, find the length of each side. For each corner, use the dot product formula with the two vectors that meet at that corner to find its cosine, and then use the inverse cosine to get the angle. Make sure the vectors point either both away or both towards the vertex for the interior angle.

🎯 Exam Tip: When finding angles of a triangle, it's critical to use vectors originating from the common vertex (e.g., for angle A, use \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)). If you use \( \overrightarrow{AB} \) and \( \overrightarrow{CA} \), you will find the external angle. Also, check if the sum of angles is \( 180^\circ \) as a verification step.

Question 10. Find the projection of:
(i) \( \vec{a} \) in the direction of \( \vec{b} \) if \( \vec{a} = 2 \hat{i} + 3 \hat{j} \), \( \vec{b} = 3 \hat{i} + 2 \hat{j} \).
(ii) \( \vec{b} \) in the direction of \( \vec{a} \) if \( \vec{a} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k} \), \( \vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} \).
Answer:
(i) Given \( \vec{a} = 2 \hat{i} + 3 \hat{j} \) and \( \vec{b} = 3 \hat{i} + 2 \hat{j} \).
The projection of \( \vec{a} \) on \( \vec{b} \) is found using the formula \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \).
First, calculate the dot product:
\( \vec{a} \cdot \vec{b} = (2)(3) + (3)(2) = 6 + 6 = 12 \)
Next, find the magnitude of \( \vec{b} \):
\( |\vec{b}| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \)
So, the projection is \( \frac{12}{\sqrt{13}} \). This value tells us how much of vector 'a' points in the same direction as vector 'b'.
In simple words: Imagine shining a light from above vector 'a' onto vector 'b'. The shadow that 'a' makes on 'b' is its projection. Here, the shadow's length is 12 divided by the square root of 13.
(ii) Given \( \vec{a} = 2 \hat{i} + 3 \hat{j} - 2 \hat{k} \) and \( \vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k} \).
The projection of \( \vec{b} \) in the direction of \( \vec{a} \) is \( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} \).
First, calculate the dot product \( \vec{b} \cdot \vec{a} \):
\( \vec{b} \cdot \vec{a} = (1)(2) + (2)(3) + (3)(-2) = 2 + 6 - 6 = 2 \)
Next, find the magnitude of \( \vec{a} \):
\( |\vec{a}| = \sqrt{2^2 + 3^2 + (-2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \)
So, the projection is \( \frac{2}{\sqrt{17}} \). This represents the component of vector 'b' that lies along the direction of vector 'a'.
In simple words: We want to find how much of vector 'b' is pointing in the same way as vector 'a'. We do this by multiplying their matching parts and dividing by the total length of vector 'a'. The result is a number that shows this overlap.

🎯 Exam Tip: Remember that the projection of vector X on vector Y is a scalar value (a number), while the vector projection of X on Y is itself a vector.

Question 11. Find the projection of \( \vec{b} + \vec{c} \) on \( \vec{a} \) where \( \vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k} \), \( \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} \) and \( \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} \).
Answer:
Given vectors:
\( \vec{a} = 2 \hat{i} - 2 \hat{j} + \hat{k} \)
\( \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k} \)
\( \vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} \)

First, find \( \vec{b} + \vec{c} \):
\( \vec{b} + \vec{c} = (\hat{i} + 2 \hat{j} - 2 \hat{k}) + (2 \hat{i} - \hat{j} + 4 \hat{k}) \)
\( \vec{b} + \vec{c} = (1+2) \hat{i} + (2-1) \hat{j} + (-2+4) \hat{k} \)
\( \vec{b} + \vec{c} = 3 \hat{i} + \hat{j} + 2 \hat{k} \)

Next, find the dot product of \( (\vec{b} + \vec{c}) \) with \( \vec{a} \):
\( (\vec{b} + \vec{c}) \cdot \vec{a} = (3 \hat{i} + \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} - 2 \hat{j} + \hat{k}) \)
\( = (3)(2) + (1)(-2) + (2)(1) \)
\( = 6 - 2 + 2 = 6 \)

Now, find the magnitude of \( \vec{a} \):
\( |\vec{a}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \)

Finally, the projection of \( (\vec{b} + \vec{c}) \) on \( \vec{a} \) is \( \frac{(\vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a}|} \):
\( \frac{6}{3} = 2 \)
The projection value indicates how much the combined vector \( \vec{b} + \vec{c} \) aligns with vector \( \vec{a} \).
In simple words: We first add vectors 'b' and 'c' together. Then, we find out how much of this new combined vector lies in the direction of vector 'a'. We do this by calculating their dot product and dividing by the length of vector 'a'. The answer is 2.

🎯 Exam Tip: Always calculate the resultant vector \( (\vec{b} + \vec{c}) \) first before finding its projection on another vector. Ensure to distinguish between scalar projection and vector projection in your answers.

Question 12. Using vectors, find the projection of the line joining the points (1, -1, 1) and (-2, 3, 4) on the line joining the points (4, -3, 5) and (7, 8, 7).
Answer:
Let the first line segment join points \( A(1, -1, 1) \) and \( B(-2, 3, 4) \).
The vector representing this line segment is \( \overrightarrow{AB} \):
\( \overrightarrow{AB} = \text{Position Vector of B} - \text{Position Vector of A} \)
\( \overrightarrow{AB} = (-2 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (\hat{i} - \hat{j} + \hat{k}) \)
\( \overrightarrow{AB} = (-2-1) \hat{i} + (3-(-1)) \hat{j} + (4-1) \hat{k} \)
\( \overrightarrow{AB} = -3 \hat{i} + 4 \hat{j} + 3 \hat{k} \)

Let the second line segment join points \( C(4, -3, 5) \) and \( D(7, 8, 7) \).
The vector representing this line segment is \( \overrightarrow{CD} \):
\( \overrightarrow{CD} = \text{Position Vector of D} - \text{Position Vector of C} \)
\( \overrightarrow{CD} = (7 \hat{i} + 8 \hat{j} + 7 \hat{k}) - (4 \hat{i} - 3 \hat{j} + 5 \hat{k}) \)
\( \overrightarrow{CD} = (7-4) \hat{i} + (8-(-3)) \hat{j} + (7-5) \hat{k} \)
\( \overrightarrow{CD} = 3 \hat{i} + 11 \hat{j} + 2 \hat{k} \)

Now we need to find the projection of \( \overrightarrow{AB} \) on \( \overrightarrow{CD} \). The formula for this is \( \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|} \).
First, calculate the dot product \( \overrightarrow{AB} \cdot \overrightarrow{CD} \):
\( \overrightarrow{AB} \cdot \overrightarrow{CD} = (-3 \hat{i} + 4 \hat{j} + 3 \hat{k}) \cdot (3 \hat{i} + 11 \hat{j} + 2 \hat{k}) \)
\( = (-3)(3) + (4)(11) + (3)(2) \)
\( = -9 + 44 + 6 = 41 \)

Next, find the magnitude of \( \overrightarrow{CD} \):
\( |\overrightarrow{CD}| = \sqrt{3^2 + 11^2 + 2^2} = \sqrt{9 + 121 + 4} = \sqrt{134} \)

Finally, the required projection is \( \frac{41}{\sqrt{134}} \). This value tells us the length of the shadow cast by the first line segment onto the second one.
In simple words: We turn the two line segments into vectors. Then, we find out how much the first vector 'points' in the same direction as the second vector. We do this by using a special calculation that combines their parts and divides by the length of the second vector.

🎯 Exam Tip: When finding a vector between two points, subtract the position vector of the starting point from the position vector of the ending point. Ensure correct calculation of magnitudes and dot products.

Question 13. Find the vectors projection of the vector \( 7 \hat{i} + \hat{j} - 4 \hat{k} \) on
(i) \( 2 \hat{i} + 6 \hat{j} + 3 \hat{k} \)
(ii) \( 7 \hat{i} + \hat{j} - 3 \hat{k} \).
Answer:
Let \( \vec{a} = 7 \hat{i} + \hat{j} - 4 \hat{k} \).

(i) Projection on \( \vec{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k} \)
The vector projection of \( \vec{a} \) on \( \vec{b} \) is given by the formula \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \).
First, calculate the dot product \( \vec{a} \cdot \vec{b} \):
\( \vec{a} \cdot \vec{b} = (7 \hat{i} + \hat{j} - 4 \hat{k}) \cdot (2 \hat{i} + 6 \hat{j} + 3 \hat{k}) \)
\( = (7)(2) + (1)(6) + (-4)(3) \)
\( = 14 + 6 - 12 = 8 \)

Next, find the magnitude of \( \vec{b} \):
\( |\vec{b}| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \)
So, \( |\vec{b}|^2 = 7^2 = 49 \).

Now, substitute these values into the projection formula:
Vector projection \( = \frac{8}{49} (2 \hat{i} + 6 \hat{j} + 3 \hat{k}) \)
\( = \frac{16}{49} \hat{i} + \frac{48}{49} \hat{j} + \frac{24}{49} \hat{k} \)
This is the part of vector 'a' that points in the exact direction of vector 'b'.
In simple words (i): We are finding the part of vector 'a' that goes exactly along vector 'b'. We calculate how much they overlap and then scale vector 'b' by that amount.

(ii) Projection on \( \vec{b'} = 7 \hat{i} + \hat{j} - 3 \hat{k} \)
Let the second vector be \( \vec{b'} = 7 \hat{i} + \hat{j} - 3 \hat{k} \).
The vector projection of \( \vec{a} \) on \( \vec{b'} \) is \( \frac{\vec{a} \cdot \vec{b'}}{|\vec{b'}|^2} \vec{b'} \).
First, calculate the dot product \( \vec{a} \cdot \vec{b'} \):
\( \vec{a} \cdot \vec{b'} = (7 \hat{i} + \hat{j} - 4 \hat{k}) \cdot (7 \hat{i} + \hat{j} - 3 \hat{k}) \)
\( = (7)(7) + (1)(1) + (-4)(-3) \)
\( = 49 + 1 + 12 = 62 \)

Next, find the magnitude of \( \vec{b'} \):
\( |\vec{b'}| = \sqrt{7^2 + 1^2 + (-3)^2} = \sqrt{49 + 1 + 9} = \sqrt{59} \)
So, \( |\vec{b'}|^2 = 59 \).

Now, substitute these values into the projection formula:
Vector projection \( = \frac{62}{59} (7 \hat{i} + \hat{j} - 3 \hat{k}) \)
This is the vector component of 'a' that aligns with 'b''.
In simple words (ii): Similar to the first part, we are finding the vector part of 'a' that points in the same direction as the new vector 'b''. We calculate their dot product and divide by the square of 'b''s length, then multiply by vector 'b'' itself.

🎯 Exam Tip: Distinguish between scalar projection \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \) (a number) and vector projection \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \) (a vector). Make sure to write the final answer in vector form for vector projection questions.

Question 14. Let \( \vec{u}, \vec{v} \) and \( \vec{w} \) be vectors such that \( \vec{u} + \vec{v} + \vec{w} = \overrightarrow{0} \). If \( |\vec{u}| = 3, |\vec{v}| = 4 \) and \( |\vec{w}| = 5 \), then find \( \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} \).
Answer:
We are given that \( \vec{u} + \vec{v} + \vec{w} = \overrightarrow{0} \).
We are also given the magnitudes: \( |\vec{u}| = 3 \), \( |\vec{v}| = 4 \), \( |\vec{w}| = 5 \).

To find \( \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} \), we can use the property of squaring the sum of vectors:
\( (\vec{u} + \vec{v} + \vec{w})^2 = \vec{u}^2 + \vec{v}^2 + \vec{w}^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \)
Since \( \vec{u} + \vec{v} + \vec{w} = \overrightarrow{0} \), then \( (\vec{u} + \vec{v} + \vec{w})^2 = 0 \).
Also, \( \vec{x}^2 = |\vec{x}|^2 \). So, \( \vec{u}^2 = |\vec{u}|^2 \), \( \vec{v}^2 = |\vec{v}|^2 \), \( \vec{w}^2 = |\vec{w}|^2 \).

Substitute these into the equation:
\( 0 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \)
Now, plug in the given magnitudes:
\( 0 = (3)^2 + (4)^2 + (5)^2 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \)
\( 0 = 9 + 16 + 25 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \)
\( 0 = 50 + 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \)
Subtract 50 from both sides:
\( -50 = 2(\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}) \)
Divide by 2:
\( \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u} = -25 \)
This result shows a relationship between the magnitudes of the vectors and their dot products when their sum is zero.
In simple words: We are given three vectors that add up to zero. We also know their individual lengths. To find the sum of their dot products, we use a special math rule where we square the sum of the vectors. Since their sum is zero, the squared sum is also zero. This helps us find the answer.

🎯 Exam Tip: This identity \( (\vec{a}+\vec{b}+\vec{c})^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) \) is crucial for solving problems involving sums of vectors and their dot products.

Question 15. Find the angles which the vector \( \vec{a} = \hat{i} - \hat{j} + \sqrt{2} \hat{k} \) makes with the coordinate axes.
Answer:
Let the angles made by the vector \( \vec{a} \) with the x-axis, y-axis, and z-axis be \( \alpha, \beta, \) and \( \gamma \) respectively. These are called direction angles.
The direction cosines are given by:
\( \cos \alpha = \frac{\vec{a} \cdot \hat{i}}{|\vec{a}|} \)
\( \cos \beta = \frac{\vec{a} \cdot \hat{j}}{|\vec{a}|} \)
\( \cos \gamma = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}|} \)

Given vector \( \vec{a} = \hat{i} - \hat{j} + \sqrt{2} \hat{k} \).
The components are \( a_x = 1 \), \( a_y = -1 \), \( a_z = \sqrt{2} \).

First, find the magnitude of \( \vec{a} \):
\( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{1^2 + (-1)^2 + (\sqrt{2})^2} \)
\( = \sqrt{1 + 1 + 2} = \sqrt{4} = 2 \)

Now, calculate the direction cosines:
For \( \alpha \) (with x-axis):
\( \cos \alpha = \frac{1}{2} \)
\( \implies \alpha = \cos^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{3} \) or \( 60^\circ \).

For \( \beta \) (with y-axis):
\( \cos \beta = \frac{-1}{2} \)
\( \implies \beta = \cos^{-1} \left( \frac{-1}{2} \right) = \frac{2\pi}{3} \) or \( 120^\circ \).

For \( \gamma \) (with z-axis):
\( \cos \gamma = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)
\( \implies \gamma = \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} \) or \( 45^\circ \).
These angles indicate the orientation of the vector in three-dimensional space.
In simple words: We want to find how much our vector tilts away from the 'x' line, the 'y' line, and the 'z' line. We first find the total length of the vector. Then, we divide each part of the vector by its total length to get the 'cosine' of each angle. From these cosine values, we can find the actual angles.

🎯 Exam Tip: The sum of the squares of direction cosines is always 1 (\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)). Use this as a quick check for your calculated direction cosines.

Question 16. Dot product of a vector with \( \vec{a_1} = \hat{i} + \hat{j} + 3 \hat{k} \), \( \vec{a_2} = \hat{i} + 3 \hat{j} - 2 \hat{k} \) and \( \vec{a_3} = 2 \hat{i} + \hat{j} + 4 \hat{k} \) are 0, 5 and 8 respectively. Find the vector.
Answer:
Let the required vector be \( \vec{x} = x \hat{i} + y \hat{j} + z \hat{k} \).
We are given three conditions based on the dot product of \( \vec{x} \) with three other vectors.

Condition 1: Dot product of \( \vec{x} \) with \( \vec{a_1} = \hat{i} + \hat{j} - 3 \hat{k} \) is 0. (Note: The solution uses \( \hat{i} + \hat{j} - 3 \hat{k} \).)
\( \vec{x} \cdot (\hat{i} + \hat{j} - 3 \hat{k}) = 0 \)
\( (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} - 3 \hat{k}) = 0 \)
\( x(1) + y(1) + z(-3) = 0 \)
\( \implies x + y - 3z = 0 \) (Equation 1)

Condition 2: Dot product of \( \vec{x} \) with \( \vec{a_2} = \hat{i} + 3 \hat{j} - 2 \hat{k} \) is 5.
\( \vec{x} \cdot (\hat{i} + 3 \hat{j} - 2 \hat{k}) = 5 \)
\( (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + 3 \hat{j} - 2 \hat{k}) = 5 \)
\( x(1) + y(3) + z(-2) = 5 \)
\( \implies x + 3y - 2z = 5 \) (Equation 2)

Condition 3: Dot product of \( \vec{x} \) with \( \vec{a_3} = 2 \hat{i} + \hat{j} + 4 \hat{k} \) is 8.
\( \vec{x} \cdot (2 \hat{i} + \hat{j} + 4 \hat{k}) = 8 \)
\( (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (2 \hat{i} + \hat{j} + 4 \hat{k}) = 8 \)
\( x(2) + y(1) + z(4) = 8 \)
\( \implies 2x + y + 4z = 8 \) (Equation 3)

Now we have a system of three linear equations:
1. \( x + y - 3z = 0 \)
2. \( x + 3y - 2z = 5 \)
3. \( 2x + y + 4z = 8 \)

Subtract Equation 1 from Equation 2:
\( (x + 3y - 2z) - (x + y - 3z) = 5 - 0 \)
\( x + 3y - 2z - x - y + 3z = 5 \)
\( \implies 2y + z = 5 \) (Equation 4)

From Equation 1, \( x = 3z - y \). Substitute this into Equation 3:
\( 2(3z - y) + y + 4z = 8 \)
\( 6z - 2y + y + 4z = 8 \)
\( 10z - y = 8 \)
\( \implies y = 10z - 8 \) (Equation 5)

Substitute Equation 5 into Equation 4:
\( 2(10z - 8) + z = 5 \)
\( 20z - 16 + z = 5 \)
\( 21z = 21 \)
\( \implies z = 1 \)

Now find \( y \) using Equation 5:
\( y = 10(1) - 8 = 10 - 8 = 2 \)

Now find \( x \) using Equation 1:
\( x + 2 - 3(1) = 0 \)
\( x + 2 - 3 = 0 \)
\( x - 1 = 0 \)
\( \implies x = 1 \)

So, the required vector is \( \vec{x} = \hat{i} + 2 \hat{j} + \hat{k} \). This vector satisfies all the given dot product conditions.
In simple words: We are looking for a mystery vector. We know how this vector behaves when multiplied (dot product) with three other known vectors. Each of these multiplications gives us a clue (an equation). By solving these three clues together, we can figure out the exact numbers that make up our mystery vector.

🎯 Exam Tip: When setting up a system of linear equations from vector dot products, double-check your signs, especially when terms are subtracted or when a component is negative. Solving these systems carefully prevents errors.

Question 17. If the sum of two unit vectors is a unit vector, prove that the magnitude of their difference is \( \sqrt{3} \).
Answer:
Let \( \vec{a} \) and \( \vec{b} \) be two unit vectors.
This means their magnitudes are 1: \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \).
We are given that their sum is also a unit vector, so \( |\vec{a} + \vec{b}| = 1 \).

We need to prove that \( |\vec{a} - \vec{b}| = \sqrt{3} \).

We use the identity for magnitudes of vector sums and differences:
\( |\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2) \)

Substitute the given values into this identity:
\( (1)^2 + |\vec{a} - \vec{b}|^2 = 2((1)^2 + (1)^2) \)
\( 1 + |\vec{a} - \vec{b}|^2 = 2(1 + 1) \)
\( 1 + |\vec{a} - \vec{b}|^2 = 2(2) \)
\( 1 + |\vec{a} - \vec{b}|^2 = 4 \)
Subtract 1 from both sides:
\( |\vec{a} - \vec{b}|^2 = 4 - 1 \)
\( |\vec{a} - \vec{b}|^2 = 3 \)
Take the square root of both sides:
\( |\vec{a} - \vec{b}| = \sqrt{3} \)
Thus, we have proved that the magnitude of their difference is \( \sqrt{3} \). This property highlights a unique geometric relationship between unit vectors under these conditions.
In simple words: We have two special vectors, each with a length of 1. When we add them, the new vector also has a length of 1. We then use a math rule that links the length of their sum to the length of their difference. By putting in the given lengths, we find that the length of their difference must be the square root of 3.

🎯 Exam Tip: The identity \( |\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2) \) is very useful for problems involving sums and differences of vector magnitudes. Memorize it and know when to apply it.

Question 18. If a and b are unit vectors inclined at the angle \( \theta \), then prove that
(i) \( \cos \frac{\theta}{2} = \frac{1}{2}|\hat{a} + \hat{b}| \),
(ii) \( \tan \frac{\theta}{2} = \frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|} \)
Answer:
Given that \( \hat{a} \) and \( \hat{b} \) are unit vectors, which means \( |\hat{a}| = 1 \) and \( |\hat{b}| = 1 \).
The angle between them is \( \theta \).
The dot product of two vectors is given by \( \hat{a} \cdot \hat{b} = |\hat{a}| |\hat{b}| \cos \theta = (1)(1) \cos \theta = \cos \theta \).

(i) Proving \( \cos \frac{\theta}{2} = \frac{1}{2}|\hat{a} + \hat{b}| \)
Consider \( |\hat{a} + \hat{b}|^2 \):
\( |\hat{a} + \hat{b}|^2 = (\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b}) \)
\( = \hat{a} \cdot \hat{a} + \hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{a} + \hat{b} \cdot \hat{b} \)
\( = |\hat{a}|^2 + |\hat{b}|^2 + 2(\hat{a} \cdot \hat{b}) \)
Substitute \( |\hat{a}| = 1 \), \( |\hat{b}| = 1 \), and \( \hat{a} \cdot \hat{b} = \cos \theta \):
\( |\hat{a} + \hat{b}|^2 = 1^2 + 1^2 + 2 \cos \theta \)
\( = 1 + 1 + 2 \cos \theta \)
\( = 2 + 2 \cos \theta \)
Factor out 2:
\( = 2(1 + \cos \theta) \)
Using the trigonometric identity \( 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2} \):
\( |\hat{a} + \hat{b}|^2 = 2(2 \cos^2 \frac{\theta}{2}) \)
\( = 4 \cos^2 \frac{\theta}{2} \)
Take the square root of both sides:
\( |\hat{a} + \hat{b}| = \sqrt{4 \cos^2 \frac{\theta}{2}} = 2 |\cos \frac{\theta}{2}| \)
Assuming \( \theta \) is in the range \( [0, \pi] \), \( \cos \frac{\theta}{2} \ge 0 \).
So, \( |\hat{a} + \hat{b}| = 2 \cos \frac{\theta}{2} \)
\( \implies \cos \frac{\theta}{2} = \frac{1}{2} |\hat{a} + \hat{b}| \)
This proves the first part of the statement.

(ii) Proving \( \tan \frac{\theta}{2} = \frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|} \)
First, consider \( |\hat{a} - \hat{b}|^2 \):
\( |\hat{a} - \hat{b}|^2 = (\hat{a} - \hat{b}) \cdot (\hat{a} - \hat{b}) \)
\( = \hat{a} \cdot \hat{a} - \hat{a} \cdot \hat{b} - \hat{b} \cdot \hat{a} + \hat{b} \cdot \hat{b} \)
\( = |\hat{a}|^2 + |\hat{b}|^2 - 2(\hat{a} \cdot \hat{b}) \)
Substitute \( |\hat{a}| = 1 \), \( |\hat{b}| = 1 \), and \( \hat{a} \cdot \hat{b} = \cos \theta \):
\( |\hat{a} - \hat{b}|^2 = 1^2 + 1^2 - 2 \cos \theta \)
\( = 2 - 2 \cos \theta \)
Factor out 2:
\( = 2(1 - \cos \theta) \)
Using the trigonometric identity \( 1 - \cos \theta = 2 \sin^2 \frac{\theta}{2} \):
\( |\hat{a} - \hat{b}|^2 = 2(2 \sin^2 \frac{\theta}{2}) \)
\( = 4 \sin^2 \frac{\theta}{2} \)
Take the square root of both sides:
\( |\hat{a} - \hat{b}| = \sqrt{4 \sin^2 \frac{\theta}{2}} = 2 |\sin \frac{\theta}{2}| \)
Assuming \( \theta \) is in the range \( [0, \pi] \), \( \sin \frac{\theta}{2} \ge 0 \).
So, \( |\hat{a} - \hat{b}| = 2 \sin \frac{\theta}{2} \)

Now, divide the expression for \( |\hat{a} - \hat{b}| \) by the expression for \( |\hat{a} + \hat{b}| \) from part (i):
\( \frac{|\hat{a} - \hat{b}|}{|\hat{a} + \hat{b}|} = \frac{2 \sin \frac{\theta}{2}}{2 \cos \frac{\theta}{2}} \)
\( \frac{|\hat{a} - \hat{b}|}{|\hat{a} + \hat{b}|} = \tan \frac{\theta}{2} \)
This proves the second part of the statement. These identities are useful in simplifying vector expressions involving angles.
In simple words: We have two vectors, each with a length of 1, and they are spread out by an angle called 'theta'. We show how the 'cosine' of half that angle is related to the length of their sum, and how the 'tangent' of half that angle is related to the lengths of both their sum and their difference. We use some special math rules for squares and angles to prove these relationships.

🎯 Exam Tip: Familiarize yourself with trigonometric half-angle identities \( 1+\cos\theta = 2\cos^2(\theta/2) \) and \( 1-\cos\theta = 2\sin^2(\theta/2) \). These are frequently used when connecting vector magnitudes to angles.

Question 19. If \( \vec{a}, \vec{b}, \vec{c} \) are three mutually perpendicular unit vectors, then prove that \( |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3} \).
Answer:
Given that \( \vec{a}, \vec{b}, \vec{c} \) are three mutually perpendicular unit vectors.
"Unit vectors" means their magnitudes are 1:
\( |\vec{a}| = 1 \), \( |\vec{b}| = 1 \), \( |\vec{c}| = 1 \).

"Mutually perpendicular" means the dot product of any two distinct vectors is 0:
\( \vec{a} \cdot \vec{b} = 0 \)
\( \vec{b} \cdot \vec{c} = 0 \)
\( \vec{c} \cdot \vec{a} = 0 \)

We need to prove that \( |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3} \).
Consider the square of the magnitude of the sum of the vectors:
\( |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \)
Expanding this dot product:
\( |\vec{a} + \vec{b} + \vec{c}|^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \)

We know that \( \vec{x} \cdot \vec{x} = |\vec{x}|^2 \). So, \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 \), \( \vec{b} \cdot \vec{b} = |\vec{b}|^2 \), \( \vec{c} \cdot \vec{c} = |\vec{c}|^2 \).
Substitute the given conditions:
\( |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \)
\( |\vec{a} + \vec{b} + \vec{c}|^2 = (1)^2 + (1)^2 + (1)^2 + 2(0 + 0 + 0) \)
\( |\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 1 + 1 + 0 \)
\( |\vec{a} + \vec{b} + \vec{c}|^2 = 3 \)

Take the square root of both sides:
\( |\vec{a} + \vec{b} + \vec{c}| = \sqrt{3} \)
This proves the statement. The result is intuitive because these vectors can be thought of as forming the sides of a cube from the origin, where the diagonal would have a length of \( \sqrt{1^2+1^2+1^2} \).
In simple words: We have three vectors, each of length 1, and they are all at right angles to each other (like the corners of a room). We want to find the length of the vector formed by adding all three together. By using a math rule that expands the square of their sum, and knowing that their lengths are 1 and they don't overlap (dot product is zero), we find that the length of their sum is the square root of 3.

🎯 Exam Tip: This result is a fundamental property of orthonormal vectors. It can be generalized for 'n' mutually perpendicular unit vectors, where the magnitude of their sum would be \( \sqrt{n} \).

Question 20. If \( |\vec{a} + \vec{b}| = 60 \), \( |\vec{a} - \vec{b}| = 40 \) and \( |\vec{b}| = 46 \), find \( |\vec{a}| \).
Answer:
We are given the following magnitudes:
\( |\vec{a} + \vec{b}| = 60 \)
\( |\vec{a} - \vec{b}| = 40 \)
\( |\vec{b}| = 46 \)

We need to find \( |\vec{a}| \).

We use the vector identity that relates the magnitudes of sum and difference:
\( |\vec{a} + \vec{b}|^2 + |\vec{a} - \vec{b}|^2 = 2(|\vec{a}|^2 + |\vec{b}|^2) \)

Substitute the given values into this identity:
\( (60)^2 + (40)^2 = 2(|\vec{a}|^2 + (46)^2) \)
\( 3600 + 1600 = 2(|\vec{a}|^2 + 2116) \)
\( 5200 = 2(|\vec{a}|^2 + 2116) \)

Divide both sides by 2:
\( \frac{5200}{2} = |\vec{a}|^2 + 2116 \)
\( 2600 = |\vec{a}|^2 + 2116 \)

Subtract 2116 from both sides to find \( |\vec{a}|^2 \):
\( |\vec{a}|^2 = 2600 - 2116 \)
\( |\vec{a}|^2 = 484 \)

Take the square root of both sides:
\( |\vec{a}| = \sqrt{484} \)
\( |\vec{a}| = 22 \)
Thus, the magnitude of vector \( \vec{a} \) is 22. This identity allows us to find unknown vector magnitudes when others are known.
In simple words: We have two vectors, 'a' and 'b'. We know the length of 'b', and also the length of what happens when we add 'a' and 'b', and when we subtract 'a' and 'b'. We use a special math rule that connects all these lengths. By plugging in the numbers we know, we can work backward to find the length of vector 'a'.

🎯 Exam Tip: This problem is a direct application of the parallelogram law of vector addition, which states that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its four sides.

Question 21. If a vector \( \vec{a} \) is perpendicular to two non-collinear vectors \( \vec{b} \) and \( \vec{c} \), then \( \vec{a} \) is perpendicular to every vector in the plane of \( \vec{b} \) and \( \vec{c} \).
Answer:
Given that vector \( \vec{a} \) is perpendicular to two non-collinear vectors \( \vec{b} \) and \( \vec{c} \).
"Perpendicular" means their dot product is 0:
\( \vec{a} \cdot \vec{b} = 0 \)
\( \vec{a} \cdot \vec{c} = 0 \)

We need to show that \( \vec{a} \) is perpendicular to every vector in the plane of \( \vec{b} \) and \( \vec{c} \).
Any vector \( \vec{r} \) in the plane of two non-collinear vectors \( \vec{b} \) and \( \vec{c} \) can be written as a linear combination of \( \vec{b} \) and \( \vec{c} \).
So, \( \vec{r} = x \vec{b} + y \vec{c} \), where \( x \) and \( y \) are scalar numbers.

Now, let's find the dot product of \( \vec{a} \) with this general vector \( \vec{r} \):
\( \vec{a} \cdot \vec{r} = \vec{a} \cdot (x \vec{b} + y \vec{c}) \)
Using the distributive property of dot products:
\( \vec{a} \cdot \vec{r} = x (\vec{a} \cdot \vec{b}) + y (\vec{a} \cdot \vec{c}) \)

Substitute the given conditions \( \vec{a} \cdot \vec{b} = 0 \) and \( \vec{a} \cdot \vec{c} = 0 \):
\( \vec{a} \cdot \vec{r} = x (0) + y (0) \)
\( \vec{a} \cdot \vec{r} = 0 + 0 \)
\( \vec{a} \cdot \vec{r} = 0 \)

Since the dot product \( \vec{a} \cdot \vec{r} = 0 \), it means that \( \vec{a} \) is perpendicular to \( \vec{r} \).
Since \( \vec{r} \) represents any arbitrary vector in the plane of \( \vec{b} \) and \( \vec{c} \), this proves that \( \vec{a} \) is perpendicular to every vector in the plane of \( \vec{b} \) and \( \vec{c} \). This is a fundamental property in vector algebra, showing how a vector normal to two basis vectors is normal to their entire span.
In simple words: Imagine vector 'a' is like a flagpole standing straight up from a flat table. On this table are two other vectors, 'b' and 'c', which point in different directions. If the flagpole 'a' is straight up from both 'b' and 'c' separately, it means the flagpole 'a' is also straight up from the entire surface of the table (the plane that 'b' and 'c' create).

🎯 Exam Tip: This concept forms the basis of finding a normal vector to a plane. Any vector perpendicular to two non-collinear vectors in a plane is perpendicular to the entire plane.

Question 22. If \( \vec{A} = \hat{i} + 2 \hat{j} + 3 \hat{k} \), \( \vec{B} = -\hat{i} + 2 \hat{j} + \hat{k} \) and \( \vec{C} = 3 \hat{i} + \hat{j} \), find t such that \( \vec{A} + t \vec{B} \) is perpendicular to \( \vec{C} \).
Answer:
Given vectors:
\( \vec{A} = \hat{i} + 2 \hat{j} + 3 \hat{k} \)
\( \vec{B} = -\hat{i} + 2 \hat{j} + \hat{k} \)
\( \vec{C} = 3 \hat{i} + \hat{j} \)

First, find the vector \( \vec{A} + t \vec{B} \):
\( \vec{A} + t \vec{B} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + t(-\hat{i} + 2 \hat{j} + \hat{k}) \)
\( = (1 - t) \hat{i} + (2 + 2t) \hat{j} + (3 + t) \hat{k} \)

We are given that \( \vec{A} + t \vec{B} \) is perpendicular to \( \vec{C} \).
If two vectors are perpendicular, their dot product is 0.
So, \( (\vec{A} + t \vec{B}) \cdot \vec{C} = 0 \).

Substitute the expressions for \( (\vec{A} + t \vec{B}) \) and \( \vec{C} \):
\( [(1 - t) \hat{i} + (2 + 2t) \hat{j} + (3 + t) \hat{k}] \cdot (3 \hat{i} + \hat{j} + 0 \hat{k}) = 0 \)
\( (1 - t)(3) + (2 + 2t)(1) + (3 + t)(0) = 0 \)
\( 3 - 3t + 2 + 2t + 0 = 0 \)
Combine like terms:
\( (3 + 2) + (-3t + 2t) = 0 \)
\( 5 - t = 0 \)
\( \implies t = 5 \)
Thus, when \( t = 5 \), the vector \( \vec{A} + t \vec{B} \) will be perpendicular to \( \vec{C} \). This calculation helps determine specific conditions for vector relationships.
In simple words: We have three vectors. We want to find a number 't' that makes a new vector (formed by adding vector 'A' to 't' times vector 'B') stand perfectly straight (perpendicular) to vector 'C'. We use the rule that if two vectors are perpendicular, their dot product is zero, and solve for 't'.

🎯 Exam Tip: Remember that two vectors are perpendicular if and only if their dot product is zero. Carefully form the resultant vector \( \vec{A} + t \vec{B} \) before taking the dot product with \( \vec{C} \).

Question 23. If A, B, C, D are any four points, show that \( \overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD} = 0 \).
Answer:
Let \( \vec{a}, \vec{b}, \vec{c}, \vec{d} \) be the position vectors of points A, B, C, D respectively, with O as the origin.

Then the vectors representing the sides are:
\( \overrightarrow{AB} = \vec{b} - \vec{a} \)
\( \overrightarrow{CD} = \vec{d} - \vec{c} \)
\( \overrightarrow{BC} = \vec{c} - \vec{b} \)
\( \overrightarrow{AD} = \vec{d} - \vec{a} \)
\( \overrightarrow{CA} = \vec{a} - \vec{c} \)
\( \overrightarrow{BD} = \vec{d} - \vec{b} \)

Now, substitute these into the given expression:
L.H.S. \( = \overrightarrow{AB} \cdot \overrightarrow{CD} + \overrightarrow{BC} \cdot \overrightarrow{AD} + \overrightarrow{CA} \cdot \overrightarrow{BD} \)
\( = (\vec{b} - \vec{a}) \cdot (\vec{d} - \vec{c}) + (\vec{c} - \vec{b}) \cdot (\vec{d} - \vec{a}) + (\vec{a} - \vec{c}) \cdot (\vec{d} - \vec{b}) \)

Expand each dot product:
1. \( (\vec{b} - \vec{a}) \cdot (\vec{d} - \vec{c}) = \vec{b} \cdot \vec{d} - \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{d} + \vec{a} \cdot \vec{c} \)
2. \( (\vec{c} - \vec{b}) \cdot (\vec{d} - \vec{a}) = \vec{c} \cdot \vec{d} - \vec{c} \cdot \vec{a} - \vec{b} \cdot \vec{d} + \vec{b} \cdot \vec{a} \)
3. \( (\vec{a} - \vec{c}) \cdot (\vec{d} - \vec{b}) = \vec{a} \cdot \vec{d} - \vec{a} \cdot \vec{b} - \vec{c} \cdot \vec{d} + \vec{c} \cdot \vec{b} \)

Now, sum these three expanded expressions:
\( (\vec{b} \cdot \vec{d} - \vec{b} \cdot \vec{c} - \vec{a} \cdot \vec{d} + \vec{a} \cdot \vec{c}) \)
\( + (\vec{c} \cdot \vec{d} - \vec{c} \cdot \vec{a} - \vec{b} \cdot \vec{d} + \vec{b} \cdot \vec{a}) \)
\( + (\vec{a} \cdot \vec{d} - \vec{a} \cdot \vec{b} - \vec{c} \cdot \vec{d} + \vec{c} \cdot \vec{b}) \)

Let's group and cancel terms. Remember that \( \vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{x} \):
\( \vec{b} \cdot \vec{d} \) and \( -\vec{b} \cdot \vec{d} \) cancel.
\( -\vec{b} \cdot \vec{c} \) and \( \vec{c} \cdot \vec{b} \) cancel.
\( -\vec{a} \cdot \vec{d} \) and \( \vec{a} \cdot \vec{d} \) cancel.
\( \vec{a} \cdot \vec{c} \) and \( -\vec{c} \cdot \vec{a} \) cancel.
\( \vec{c} \cdot \vec{d} \) and \( -\vec{c} \cdot \vec{d} \) cancel.
\( \vec{b} \cdot \vec{a} \) and \( -\vec{a} \cdot \vec{b} \) cancel.

All terms cancel out, so the sum is 0.
L.H.S. = 0 = R.H.S.
Hence proved. This identity demonstrates a cyclic vector relationship for any four points, regardless of their specific arrangement.
In simple words: We are given four points and want to show a special relationship between the dot products of vectors connecting these points. We represent each vector using position vectors. Then, we write out the full expression and multiply everything inside. When we combine all the terms, we find that they all cancel each other out, proving that the total sum is zero.

🎯 Exam Tip: When proving vector identities, expressing all vectors in terms of position vectors from a common origin often simplifies the algebraic manipulation. Remember that dot product is commutative, i.e., \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \).

Question 24. If \( \vec{a} \) and \( \vec{b} \) are vectors with a, b as their respective lengths, show that \( \left(\frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2}\right)^2 = \left(\frac{\vec{a}-\vec{b}}{ab}\right)^2 \).
Answer:
Given that \( a = |\vec{a}| \) and \( b = |\vec{b}| \).
We need to show: \( \left(\frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2}\right)^2 = \left(\frac{\vec{a}-\vec{b}}{ab}\right)^2 \)

Let's expand the Left Hand Side (L.H.S.):
L.H.S. \( = \left(\frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2}\right)^2 \)
This is a dot product of a vector with itself: \( \left(\frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2}\right) \cdot \left(\frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2}\right) \)
\( = \frac{\vec{a}}{a^2} \cdot \frac{\vec{a}}{a^2} - \frac{\vec{a}}{a^2} \cdot \frac{\vec{b}}{b^2} - \frac{\vec{b}}{b^2} \cdot \frac{\vec{a}}{a^2} + \frac{\vec{b}}{b^2} \cdot \frac{\vec{b}}{b^2} \)
\( = \frac{\vec{a} \cdot \vec{a}}{a^4} - \frac{\vec{a} \cdot \vec{b}}{a^2 b^2} - \frac{\vec{b} \cdot \vec{a}}{a^2 b^2} + \frac{\vec{b} \cdot \vec{b}}{b^4} \)
Since \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 = a^2 \) and \( \vec{b} \cdot \vec{b} = |\vec{b}|^2 = b^2 \), and \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \):
\( = \frac{a^2}{a^4} - \frac{\vec{a} \cdot \vec{b}}{a^2 b^2} - \frac{\vec{a} \cdot \vec{b}}{a^2 b^2} + \frac{b^2}{b^4} \)
\( = \frac{1}{a^2} - \frac{2(\vec{a} \cdot \vec{b})}{a^2 b^2} + \frac{1}{b^2} \) (Equation 1)

Now, let's expand the Right Hand Side (R.H.S.):
R.H.S. \( = \left(\frac{\vec{a}-\vec{b}}{ab}\right)^2 \)
This is also a dot product: \( \frac{(\vec{a}-\vec{b})}{ab} \cdot \frac{(\vec{a}-\vec{b})}{ab} \)
\( = \frac{(\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b})}{a^2 b^2} \)
\( = \frac{\vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}}{a^2 b^2} \)
Using \( \vec{a} \cdot \vec{a} = a^2 \), \( \vec{b} \cdot \vec{b} = b^2 \), and \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \):
\( = \frac{a^2 - 2(\vec{a} \cdot \vec{b}) + b^2}{a^2 b^2} \)
Separate the terms:
\( = \frac{a^2}{a^2 b^2} - \frac{2(\vec{a} \cdot \vec{b})}{a^2 b^2} + \frac{b^2}{a^2 b^2} \)
\( = \frac{1}{b^2} - \frac{2(\vec{a} \cdot \vec{b})}{a^2 b^2} + \frac{1}{a^2} \) (Equation 2)

Comparing Equation 1 and Equation 2, we can see that they are identical.
L.H.S. = R.H.S.
Hence proved. This identity shows a symmetrical relationship between scaled vector differences and their magnitudes.
In simple words: We are asked to show that two complex vector expressions are equal. We take the left side and expand it using dot product rules, simplifying it based on what we know about vector lengths. Then, we do the same for the right side. After simplifying both sides, we see that they are exactly the same, which proves the statement.

🎯 Exam Tip: Remember that squaring a vector expression means taking its dot product with itself, e.g., \( \vec{X}^2 = \vec{X} \cdot \vec{X} \). Also, carefully handle the scalar terms in the denominator to avoid common errors.