GSEB Class 12 Maths Solutions Chapter 10 સદિશ બીજગણિત Exercise 10.4

Question 1. If \( \vec{a} = \hat{i} – 7\hat{j} + 7\hat{k} \) and \( \vec{b} = 3\hat{i} – 2\hat{j} + 2\hat{k} \), then find \( |\vec{a} \times \vec{b}| \).
Answer:
Given vectors are \( \vec{a} = \hat{i} – 7\hat{j} + 7\hat{k} \) and \( \vec{b} = 3\hat{i} – 2\hat{j} + 2\hat{k} \).
First, we calculate the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix} \] \( \implies \vec{a} \times \vec{b} = \hat{i} ((-7)(2) - (7)(-2)) - \hat{j} ((1)(2) - (7)(3)) + \hat{k} ((1)(-2) - (-7)(3)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (-14 + 14) - \hat{j} (2 - 21) + \hat{k} (-2 + 21) \)
\( \implies \vec{a} \times \vec{b} = 0\hat{i} - \hat{j} (-19) + \hat{k} (19) \)
\( \implies \vec{a} \times \vec{b} = 19\hat{j} + 19\hat{k} \)
Next, we find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(0)^2 + (19)^2 + (19)^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{0 + 361 + 361} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{722} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{2 \times 361} \)
\( \implies |\vec{a} \times \vec{b}| = 19\sqrt{2} \)
The magnitude of \( \vec{a} \times \vec{b} \) is \( 19\sqrt{2} \).
In simple words: First, we multiply the two given vectors using the cross product rule. This gives us a new vector. Then, we find the length of this new vector by using the square root of the sum of the squares of its components.

Exam Tip: Remember that the cross product of two vectors is a new vector perpendicular to both. Its magnitude represents the area of the parallelogram formed by the two vectors.

Question 2. If \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} – 2\hat{k} \), then find the unit vector perpendicular to \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \).
Answer:
Given vectors are \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} – 2\hat{k} \).
First, let's find \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \):
\( \vec{a}+\vec{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) + (\hat{i} + 2\hat{j} – 2\hat{k}) \)
\( \implies \vec{a}+\vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} \)
\( \implies \vec{a}+\vec{b} = 4\hat{i} + 4\hat{j} + 0\hat{k} \)
\( \vec{a}-\vec{b} = (3\hat{i} + 2\hat{j} + 2\hat{k}) - (\hat{i} + 2\hat{j} – 2\hat{k}) \)
\( \implies \vec{a}-\vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} \)
\( \implies \vec{a}-\vec{b} = 2\hat{i} + 0\hat{j} + 4\hat{k} \)
A vector perpendicular to both \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \) is given by their cross product:
\( (\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} \)
\( \implies (\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = \hat{i} ((4)(4) - (0)(0)) - \hat{j} ((4)(4) - (0)(2)) + \hat{k} ((4)(0) - (4)(2)) \)
\( \implies (\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = \hat{i} (16 - 0) - \hat{j} (16 - 0) + \hat{k} (0 - 8) \)
\( \implies (\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = 16\hat{i} - 16\hat{j} - 8\hat{k} \)
Now, we calculate the magnitude of this cross product:
\( |(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})| = \sqrt{(16)^2 + (-16)^2 + (-8)^2} \)
\( \implies |(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})| = \sqrt{256 + 256 + 64} \)
\( \implies |(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})| = \sqrt{576} \)
\( \implies |(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})| = 24 \)
The unit vector perpendicular to \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \) is given by:
\( \hat{n} = \pm \frac{(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})}{|(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})|} \)
\( \implies \hat{n} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} \)
\( \implies \hat{n} = \pm \left( \frac{16}{24}\hat{i} - \frac{16}{24}\hat{j} - \frac{8}{24}\hat{k} \right) \)
\( \implies \hat{n} = \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) \)
The unit vector perpendicular to \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \) is \( \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right) \).
In simple words: First, we add the two vectors and subtract them to get two new vectors. Then, we find a vector that is straight up from both of these new vectors using the cross product. Finally, we make this perpendicular vector into a "unit vector" by dividing it by its length, so its length becomes 1. We include both positive and negative directions because a perpendicular vector can point in two opposite ways.

Exam Tip: Remember to simplify the fractional components of the unit vector after dividing by the magnitude. The `\pm` sign is important because there are two opposite directions for a unit vector perpendicular to a plane.

Question 3. If a unit vector \( \vec{a} \) makes an angle \( \frac{\pi}{3} \) with \( \hat{i} \), \( \frac{\pi}{4} \) with \( \hat{j} \), and an acute angle \( \theta \) with \( \hat{k} \), then find \( \theta \) and the components of \( \vec{a} \).
Answer:
Let the angles made by the unit vector \( \vec{a} \) with \( \hat{i}, \hat{j}, \hat{k} \) be \( \alpha, \beta, \gamma \) respectively.
Given: \( \alpha = \frac{\pi}{3} \), \( \beta = \frac{\pi}{4} \), and \( \gamma = \theta \) (where \( \theta \) is an acute angle).
We know the direction cosines are \( \cos \alpha, \cos \beta, \cos \gamma \).
\( \cos \alpha = \cos \frac{\pi}{3} = \frac{1}{2} \)
\( \cos \beta = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
We also know that for direction cosines, the sum of their squares is always 1:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)
Substituting the known values:
\( \left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \theta = 1 \)
\( \implies \frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1 \)
\( \implies \frac{1+2}{4} + \cos^2 \theta = 1 \)
\( \implies \frac{3}{4} + \cos^2 \theta = 1 \)
\( \implies \cos^2 \theta = 1 - \frac{3}{4} \)
\( \implies \cos^2 \theta = \frac{1}{4} \)
\( \implies \cos \theta = \pm \sqrt{\frac{1}{4}} \)
\( \implies \cos \theta = \pm \frac{1}{2} \)
Since \( \theta \) is an acute angle, \( \cos \theta \) must be positive.
So, \( \cos \theta = \frac{1}{2} \)
\( \implies \theta = \frac{\pi}{3} \)
Now, let the unit vector \( \vec{a} \) be \( x\hat{i} + y\hat{j} + z\hat{k} \). Since it is a unit vector, its magnitude \( |\vec{a}| = 1 \).
The components are given by:
\( x = |\vec{a}| \cos \alpha = 1 \times \frac{1}{2} = \frac{1}{2} \)
\( y = |\vec{a}| \cos \beta = 1 \times \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \)
\( z = |\vec{a}| \cos \gamma = 1 \times \frac{1}{2} = \frac{1}{2} \)
Therefore, the vector \( \vec{a} = x\hat{i} + y\hat{j} + z\hat{k} \) is:
\( \vec{a} = \frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k} \)
The value of \( \theta \) is \( \frac{\pi}{3} \) and the components of \( \vec{a} \) are \( \frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2} \).
In simple words: We used the rule that the squares of the cosines of the angles a vector makes with the axes add up to one. We already knew two angles, so we could find the third one. Because it's a unit vector, its parts along the axes are just the cosines of those angles.

Exam Tip: Always remember the fundamental identity \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \) for direction cosines. If the angle is specified as acute, choose the positive value for cosine.

Question 4. Show that \( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \).
Answer:
We need to show that \( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \).
Let's expand the left-hand side (LHS) using the distributive property of the cross product:
\( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = \vec{a} \times (\vec{a}+\vec{b}) - \vec{b} \times (\vec{a}+\vec{b}) \)
\( \implies (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = (\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b}) - (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{b}) \)
We know that the cross product of a vector with itself is the zero vector (i.e., \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{b} = \vec{0} \)).
Also, we know that \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \).
Substituting these properties into the expression:
\( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = \vec{0} + (\vec{a} \times \vec{b}) - (-(\vec{a} \times \vec{b})) - \vec{0} \)
\( \implies (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{b}) \)
\( \implies (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \)
Thus, the left-hand side is equal to the right-hand side, and the identity is proven.
In simple words: We broke down the cross product on the left side into simpler parts. We used the rules that a vector crossed with itself is zero, and swapping vectors in a cross product changes its sign. When we put all these rules together, we got exactly what was on the right side.

Exam Tip: Remember the two key properties of vector cross products used here: \( \vec{v} \times \vec{v} = \vec{0} \) and \( \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) \). These are vital for simplifying such expressions.

Question 5. If \( (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0} \), then find \( \lambda \) and \( \mu \).
Answer:
Given the cross product is \( \vec{0} \):
\( (2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0} \)
We can write the cross product using a determinant:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = \vec{0} \] Expanding the determinant:
\( \hat{i} (6\mu - 27\lambda) - \hat{j} (2\mu - 27) + \hat{k} (2\lambda - 6) = \vec{0} \)
For this vector to be the zero vector, each of its components must be zero.
So, we get a system of equations:
1. \( 6\mu - 27\lambda = 0 \)
2. \( 2\mu - 27 = 0 \)
3. \( 2\lambda - 6 = 0 \)
From equation (3):
\( 2\lambda - 6 = 0 \)
\( \implies 2\lambda = 6 \)
\( \implies \lambda = 3 \)
From equation (2):
\( 2\mu - 27 = 0 \)
\( \implies 2\mu = 27 \)
\( \implies \mu = \frac{27}{2} \)
Let's check if these values satisfy equation (1):
\( 6\mu - 27\lambda = 6\left(\frac{27}{2}\right) - 27(3) \)
\( = 3 \times 27 - 81 \)
\( = 81 - 81 = 0 \)
The values satisfy all equations. Thus, \( \lambda = 3 \) and \( \mu = \frac{27}{2} \).
In simple words: When the cross product of two vectors is zero, it means the vectors are parallel. We set up the cross product as a determinant and made each component equal to zero. This gave us a set of equations that we solved to find the missing values.

Exam Tip: Remember that if \( \vec{A} \times \vec{B} = \vec{0} \), it implies that \( \vec{A} \) and \( \vec{B} \) are parallel vectors, meaning their components are proportional. You can use either the determinant method or the proportionality of components to solve for unknowns.

Question 6. If \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then \( \vec{a} \times \vec{b}=\vec{0} \). Is the converse true? Support your answer with an example.
Answer:
The statement "If \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then \( \vec{a} \times \vec{b}=\vec{0} \)" is true.
This is because the magnitude of the cross product is given by \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \). If either \( |\vec{a}| = 0 \) or \( |\vec{b}| = 0 \), then \( |\vec{a} \times \vec{b}| = 0 \), which means \( \vec{a} \times \vec{b}=\vec{0} \).
Now let's consider the converse: "If \( \vec{a} \times \vec{b}=\vec{0} \), then \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \)".
The converse is **not true**.
When \( \vec{a} \times \vec{b}=\vec{0} \), it means that \( |\vec{a}||\vec{b}|\sin\theta = 0 \). This can happen in three cases:
1. \( |\vec{a}| = 0 \) (i.e., \( \vec{a}=\vec{0} \))
2. \( |\vec{b}| = 0 \) (i.e., \( \vec{b}=\vec{0} \))
3. \( \sin\theta = 0 \), which implies \( \theta = 0 \) or \( \theta = \pi \). This means \( \vec{a} \) and \( \vec{b} \) are parallel or collinear vectors.
So, \( \vec{a} \times \vec{b}=\vec{0} \) also holds true if \( \vec{a} \) and \( \vec{b} \) are non-zero parallel vectors.
**Example:**
Let \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k} \).
Here, both \( \vec{a} \neq \vec{0} \) and \( \vec{b} \neq \vec{0} \).
However, \( \vec{b} = 2\vec{a} \), which means \( \vec{a} \) and \( \vec{b} \) are parallel vectors.
Let's calculate their cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{vmatrix} \] \( \implies \vec{a} \times \vec{b} = \hat{i} (1 \times 2 - 1 \times 2) - \hat{j} (1 \times 2 - 1 \times 2) + \hat{k} (1 \times 2 - 1 \times 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (0) - \hat{j} (0) + \hat{k} (0) = \vec{0} \)
In this example, \( \vec{a} \neq \vec{0} \) and \( \vec{b} \neq \vec{0} \), but \( \vec{a} \times \vec{b}=\vec{0} \). Therefore, the converse is not true.
In simple words: The first statement says if either vector is zero, their cross product is zero. This is always true. But the reverse, if their cross product is zero, does not mean one of them must be zero. It can also mean the vectors are parallel, even if they are both non-zero. Our example shows two non-zero parallel vectors whose cross product is zero.

Exam Tip: Understand that \( \vec{a} \times \vec{b} = \vec{0} \) is the condition for two vectors to be parallel or collinear, not just for one of them to be the zero vector. Providing a clear example like the one above is essential to support your answer for converse statements.

Question 7. Given vectors \( \vec{a}, \vec{b}, \vec{c} \) as \( a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \), \( b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \), and \( c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \) respectively. Prove that \( \vec{a} \times (\vec{b}+\vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \).
Answer:
Given vectors:
\( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \)
\( \vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \)
\( \vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k} \)

First, let's calculate \( \vec{b}+\vec{c} \):
\( \vec{b}+\vec{c} = (b_1+c_1)\hat{i} + (b_2+c_2)\hat{j} + (b_3+c_3)\hat{k} \)

**Left-Hand Side (LHS):** \( \vec{a} \times (\vec{b}+\vec{c}) \)
\[ \vec{a} \times (\vec{b}+\vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1+c_1 & b_2+c_2 & b_3+c_3 \end{vmatrix} \] Expanding the determinant:
\( \hat{i} [a_2(b_3+c_3) - a_3(b_2+c_2)] - \hat{j} [a_1(b_3+c_3) - a_3(b_1+c_1)] + \hat{k} [a_1(b_2+c_2) - a_2(b_1+c_1)] \)
\( \implies \hat{i} [a_2b_3+a_2c_3 - a_3b_2-a_3c_2] - \hat{j} [a_1b_3+a_1c_3 - a_3b_1-a_3c_1] + \hat{k} [a_1b_2+a_1c_2 - a_2b_1-a_2c_1] \)
Rearranging terms based on \( b \) and \( c \):
\( \implies [\hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1)] + [\hat{i}(a_2c_3 - a_3c_2) - \hat{j}(a_1c_3 - a_3c_1) + \hat{k}(a_1c_2 - a_2c_1)] \quad \ldots (1) \)

**Right-Hand Side (RHS):** \( \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \)
First, calculate \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \] \( \implies \vec{a} \times \vec{b} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1) \)
Next, calculate \( \vec{a} \times \vec{c} \):
\[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \] \( \implies \vec{a} \times \vec{c} = \hat{i}(a_2c_3 - a_3c_2) - \hat{j}(a_1c_3 - a_3c_1) + \hat{k}(a_1c_2 - a_2c_1) \)
Now, add \( \vec{a} \times \vec{b} \) and \( \vec{a} \times \vec{c} \):
\( \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = [\hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1)] + [\hat{i}(a_2c_3 - a_3c_2) - \hat{j}(a_1c_3 - a_3c_1) + \hat{k}(a_1c_2 - a_2c_1)] \quad \ldots (2) \)
From (1) and (2), we can see that LHS = RHS.
Therefore, \( \vec{a} \times (\vec{b}+\vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \).
In simple words: To prove this, we first calculated the left side by adding vectors b and c, then taking their cross product with vector a. Next, we calculated the right side by finding the cross product of a with b, then a with c, and adding those two results. Both sides came out to be exactly the same, which proves the identity.

Exam Tip: This proof demonstrates the distributive property of the vector cross product over vector addition. Breaking down the LHS and RHS and showing they are identical is the standard approach. Be careful with signs when expanding determinants.

Question 8. If \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then \( \vec{a} \times \vec{b}=\vec{0} \). Is the converse true? Support your answer with an example.
Answer:
The statement "If \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \), then \( \vec{a} \times \vec{b}=\vec{0} \)" is true.
This happens because the size of the cross product is \( |\vec{a}||\vec{b}|\sin\theta \). If either \( |\vec{a}| \) or \( |\vec{b}| \) is zero, then the whole expression becomes zero, meaning \( \vec{a} \times \vec{b}=\vec{0} \).
Now consider the reverse statement: "If \( \vec{a} \times \vec{b}=\vec{0} \), then \( \vec{a} = \vec{0} \) or \( \vec{b} = \vec{0} \)".
The reverse statement is **not true**.
When \( \vec{a} \times \vec{b}=\vec{0} \), it means that \( |\vec{a}||\vec{b}|\sin\theta = 0 \). This can occur in three situations:
1. \( |\vec{a}| = 0 \) (which means \( \vec{a}=\vec{0} \))
2. \( |\vec{b}| = 0 \) (which means \( \vec{b}=\vec{0} \))
3. \( \sin\theta = 0 \), meaning \( \theta = 0 \) or \( \theta = \pi \). This indicates that \( \vec{a} \) and \( \vec{b} \) are parallel or move along the same line.
So, \( \vec{a} \times \vec{b}=\vec{0} \) can also happen if \( \vec{a} \) and \( \vec{b} \) are non-zero but parallel vectors.
**Example:**
Let's take \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 2\hat{j} + 2\hat{k} \).
Here, both vectors are not zero: \( \vec{a} \neq \vec{0} \) and \( \vec{b} \neq \vec{0} \).
However, \( \vec{b} = 2\vec{a} \), which shows that \( \vec{a} \) and \( \vec{b} \) are parallel vectors.
Let's find their cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{vmatrix} \] \( \implies \vec{a} \times \vec{b} = \hat{i} (1 \times 2 - 1 \times 2) - \hat{j} (1 \times 2 - 1 \times 2) + \hat{k} (1 \times 2 - 1 \times 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i} (0) - \hat{j} (0) + \hat{k} (0) = \vec{0} \)
In this instance, even though \( \vec{a} \neq \vec{0} \) and \( \vec{b} \neq \vec{0} \), their cross product is \( \vec{0} \). Therefore, the converse is incorrect.
In simple words: The first part of the rule is correct: if any vector is zero, their cross product is zero. But the opposite is not always true. If the cross product is zero, it could also mean the two vectors are running in the same direction, or opposite directions, even if neither of them is zero.

Exam Tip: When evaluating converse statements, remember to look for counter-examples. The cross product being zero does not exclusively imply null vectors; it primarily indicates parallelism, which is a common area for misconceptions.

Question 9. Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
Answer:
Given the vertices of the triangle as A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).
We can find two adjacent sides of the triangle as vectors, for example, \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \).
Position vectors of the points are:
\( \overrightarrow{OA} = \hat{i} + \hat{j} + 2\hat{k} \)
\( \overrightarrow{OB} = 2\hat{i} + 3\hat{j} + 5\hat{k} \)
\( \overrightarrow{OC} = \hat{i} + 5\hat{j} + 5\hat{k} \)

Calculate \( \overrightarrow{AB} \):
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)
\( \implies \overrightarrow{AB} = (2\hat{i} + 3\hat{j} + 5\hat{k}) - (\hat{i} + \hat{j} + 2\hat{k}) \)
\( \implies \overrightarrow{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} \)
\( \implies \overrightarrow{AB} = \hat{i} + 2\hat{j} + 3\hat{k} \)

Calculate \( \overrightarrow{AC} \):
\( \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} \)
\( \implies \overrightarrow{AC} = (\hat{i} + 5\hat{j} + 5\hat{k}) - (\hat{i} + \hat{j} + 2\hat{k}) \)
\( \implies \overrightarrow{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} \)
\( \implies \overrightarrow{AC} = 0\hat{i} + 4\hat{j} + 3\hat{k} \)

Next, find the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \):
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} \] \( \implies \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}((2)(3) - (3)(4)) - \hat{j}((1)(3) - (3)(0)) + \hat{k}((1)(4) - (2)(0)) \)
\( \implies \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}(6 - 12) - \hat{j}(3 - 0) + \hat{k}(4 - 0) \)
\( \implies \overrightarrow{AB} \times \overrightarrow{AC} = -6\hat{i} - 3\hat{j} + 4\hat{k} \)

Now, find the magnitude of this cross product:
\( |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-6)^2 + (-3)^2 + (4)^2} \)
\( \implies |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{36 + 9 + 16} \)
\( \implies |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{61} \)

The area of the triangle is given by \( \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| \):
Area \( = \frac{1}{2} \sqrt{61} \) square units.
In simple words: We found two sides of the triangle as vectors. Then, we calculated their cross product and its length. The area of the triangle is half of that length.

Exam Tip: To find the area of a triangle given its vertices, first determine two vectors representing adjacent sides. Then, calculate the magnitude of their cross product and divide by two. Ensure accurate vector subtraction and determinant expansion.

Question 10. If the adjacent sides of a parallelogram are given by vectors \( \vec{a} = \hat{i} – \hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} – 7\hat{j} + \hat{k} \), find its area.
Answer:
Given the adjacent sides of a parallelogram as vectors:
\( \vec{a} = \hat{i} – \hat{j} + 3\hat{k} \)
\( \vec{b} = 2\hat{i} – 7\hat{j} + \hat{k} \)
The area of a parallelogram with adjacent sides \( \vec{a} \) and \( \vec{b} \) is given by the magnitude of their cross product, \( |\vec{a} \times \vec{b}| \).
First, let's find the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} \] \( \implies \vec{a} \times \vec{b} = \hat{i}((-1)(1) - (3)(-7)) - \hat{j}((1)(1) - (3)(2)) + \hat{k}((1)(-7) - (-1)(2)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(-1 + 21) - \hat{j}(1 - 6) + \hat{k}(-7 + 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(20) - \hat{j}(-5) + \hat{k}(-5) \)
\( \implies \vec{a} \times \vec{b} = 20\hat{i} + 5\hat{j} - 5\hat{k} \)
Now, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{(20)^2 + (5)^2 + (-5)^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{400 + 25 + 25} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{450} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{225 \times 2} \)
\( \implies |\vec{a} \times \vec{b}| = 15\sqrt{2} \)
The area of the parallelogram is \( 15\sqrt{2} \) square units.
In simple words: We find the vector that points perpendicular to both given side vectors using the cross product. Then, we measure the length of this new vector. That length is exactly the area of the parallelogram formed by the two original side vectors.

Exam Tip: Always remember that the magnitude of the cross product of two adjacent sides gives the area of the parallelogram. Ensure your determinant calculation is precise to avoid errors in the final magnitude.

Question 11. Let \( \vec{a} \) and \( \vec{b} \) be given vectors such that \( |\vec{a}| = 3 \) and \( |\vec{b}|=\frac{\sqrt{2}}{3} \). If \( \vec{a} \times \vec{b} \) is a unit vector, then the angle between \( \vec{a} \) and \( \vec{b} \) is ........
(A) \( \frac{\pi}{6} \)
(B) \( \frac{\pi}{4} \)
(C) \( \frac{\pi}{3} \)
(D) \( \frac{\pi}{2} \)
Answer: (B) \( \frac{\pi}{4} \)
In simple words: We use the formula for the magnitude of a cross product to find the angle. Since the cross product is a unit vector, its magnitude is 1. We plug in the known magnitudes of the individual vectors and solve for the sine of the angle, which helps us find the angle itself.

Exam Tip: For problems involving the angle between vectors when their cross product is known, use the formula \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \). If \( \vec{a} \times \vec{b} \) is a unit vector, its magnitude is 1.

Question 12. Position vectors of vertices A, B, C, D of a square are \( -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k} \), \( \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k} \), \( \hat{i} – \frac{1}{2}\hat{j} + 4\hat{k} \) and \( -\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k} \) respectively. Then the area of the square is...
(A) \( \frac{1}{2} \)
(B) 1
(C) 2
(D) 4
Answer: (C) 2
Let the position vectors of the vertices be:
\( \overrightarrow{OA} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k} \)
\( \overrightarrow{OB} = \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k} \)
\( \overrightarrow{OC} = \hat{i} – \frac{1}{2}\hat{j} + 4\hat{k} \)
\( \overrightarrow{OD} = -\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k} \)

First, find the vector representing side AB:
\( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} \)
\( \implies \overrightarrow{AB} = (\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}) - (-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}) \)
\( \implies \overrightarrow{AB} = (1 - (-1))\hat{i} + (\frac{1}{2} - \frac{1}{2})\hat{j} + (4 - 4)\hat{k} \)
\( \implies \overrightarrow{AB} = 2\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i} \)

Next, find the vector representing side AC:
\( \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} \)
\( \implies \overrightarrow{AC} = (\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}) - (-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}) \)
\( \implies \overrightarrow{AC} = (1 - (-1))\hat{i} + (-\frac{1}{2} - \frac{1}{2})\hat{j} + (4 - 4)\hat{k} \)
\( \implies \overrightarrow{AC} = 2\hat{i} - \hat{j} + 0\hat{k} = 2\hat{i} - \hat{j} \)

The area of the parallelogram (or rectangle, as implied by the coordinates) formed by \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) as adjacent sides is \( |\overrightarrow{AB} \times \overrightarrow{AC}| \).
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 2 & -1 & 0 \end{vmatrix} \] \( \implies \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}((0)(0) - (0)(-1)) - \hat{j}((2)(0) - (0)(2)) + \hat{k}((2)(-1) - (0)(2)) \)
\( \implies \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}(0) - \hat{j}(0) + \hat{k}(-2 - 0) \)
\( \implies \overrightarrow{AB} \times \overrightarrow{AC} = -2\hat{k} \)

The magnitude of this cross product is:
\( |\overrightarrow{AB} \times \overrightarrow{AC}| = |-2\hat{k}| = \sqrt{0^2 + 0^2 + (-2)^2} = \sqrt{4} = 2 \)
The area of the rectangle (or parallelogram) is 2 square units.
In simple words: We first found the vectors for two adjacent sides of the shape. Then, we calculated the cross product of these two vectors and found its length. This length gives us the exact area of the shape.

Exam Tip: When given coordinates, calculate the vectors for adjacent sides. The area of a parallelogram (or square/rectangle, which are special parallelograms) is the magnitude of the cross product of these adjacent side vectors. Be careful with vector subtraction for position vectors.