RBSE Solutions Class 12 Maths Chapter 13 Vector Exercise 13.3

Question 1. Find vector product of vectors \( 3\hat{i}+\hat{j}-\hat{k} \) and \( 2\hat{i}+3\hat{j}+\hat{k} \).
Answer: Let \( \vec{a} = 3\hat{i}+\hat{j}-\hat{k} \) and \( \vec{b} = 2\hat{i}+3\hat{j}+\hat{k} \).
To find the vector product (cross product) of these vectors, we set up a determinant:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -1 \\ 2 & 3 & 1 \end{vmatrix} \)
Now, we calculate the determinant:
\( \implies \vec{a} \times \vec{b} = \hat{i}(1 \times 1 - (-1) \times 3) - \hat{j}(3 \times 1 - (-1) \times 2) + \hat{k}(3 \times 3 - 1 \times 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(1 + 3) - \hat{j}(3 + 2) + \hat{k}(9 - 2) \)
\( \implies \vec{a} \times \vec{b} = 4\hat{i} - 5\hat{j} + 7\hat{k} \)
Thus, the cross product of the two given vectors is \( 4\hat{i} - 5\hat{j} + 7\hat{k} \). This new vector is perpendicular to both original vectors.
In simple words: We find the vector product by setting up a special grid called a determinant and solving it. The answer is a new vector that stands straight up from the plane of the first two vectors.

🎯 Exam Tip: Remember to correctly handle the signs when expanding the determinant for the cross product, especially the minus sign for the middle (\( \hat{j} \)) component.

Question 2. Find perpendicular unit vector of vectors \( \hat{i}-2\hat{j}+\hat{k} \) and \( 2\hat{i}+\hat{j}-3\hat{k} \).
Answer: Let \( \vec{a} = \hat{i}-2\hat{j}+\hat{k} \) and \( \vec{b} = 2\hat{i}+\hat{j}-3\hat{k} \).
First, we find the cross product of these two vectors:
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 2 & 1 & -3 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((-2) \times (-3) - 1 \times 1) - \hat{j}(1 \times (-3) - 1 \times 2) + \hat{k}(1 \times 1 - (-2) \times 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(6 - 1) - \hat{j}(-3 - 2) + \hat{k}(1 + 4) \)
\( \implies \vec{a} \times \vec{b} = 5\hat{i} + 5\hat{j} + 5\hat{k} \)
Next, we find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{5^2 + 5^2 + 5^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{25 + 25 + 25} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{75} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{25 \times 3} = 5\sqrt{3} \)
The unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) is given by \( \hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \).
\( \implies \hat{n} = \frac{5\hat{i} + 5\hat{j} + 5\hat{k}}{5\sqrt{3}} \)
\( \implies \hat{n} = \frac{5(\hat{i} + \hat{j} + \hat{k})}{5\sqrt{3}} \)
\( \implies \hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \)
This unit vector is normal (perpendicular) to both given vectors, helping us define the direction without considering length.
In simple words: First, we multiply the two vectors in a special way (cross product) to get a new vector that is perpendicular to both of them. Then, we make this new vector a "unit vector" by dividing it by its length, so it only shows the direction.

🎯 Exam Tip: To find a unit vector perpendicular to two given vectors, first calculate their cross product, and then divide the resulting vector by its magnitude.

Question 3. For vectors \( \vec{a} \) and \( \vec{b} \), prove that \( |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 \).
Answer: Let \( \theta \) be the angle between vectors \( \vec{a} \) and \( \vec{b} \).
We know the definition of the magnitude of the cross product:
\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \)
Squaring both sides, we get:
\( \implies |\vec{a} \times \vec{b}|^2 = (|\vec{a}| |\vec{b}| \sin \theta)^2 \)
\( \implies |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta \)
Using the trigonometric identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\( \implies |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 (1 - \cos^2 \theta) \)
Distribute \( |\vec{a}|^2 |\vec{b}|^2 \):
\( \implies |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta \)
We also know the definition of the dot product:
\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \)
Squaring both sides:
\( \implies (\vec{a} \cdot \vec{b})^2 = (|\vec{a}| |\vec{b}| \cos \theta)^2 \)
\( \implies (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta \)
Substitute this back into the equation for \( |\vec{a} \times \vec{b}|^2 \):
\( \implies |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 \)
This identity, also known as Lagrange's identity for vectors, is a useful tool that connects dot and cross products, enabling us to find one if the other is known.
In simple words: This rule shows a connection between the "cross product" and "dot product" of two vectors. It says that the square of the length of their cross product is equal to the product of their lengths squared, minus the square of their dot product.

🎯 Exam Tip: This identity is very important for solving problems where you are given the dot product and magnitudes and need to find the cross product magnitude, or vice versa.

Question 4. Prove that \( \vec{a} \times (\vec{b}+\vec{c}) + \vec{b} \times (\vec{c}+\vec{a}) + \vec{c} \times (\vec{a}+\vec{b}) = \vec{0} \).
Answer: We start with the Left Hand Side (L.H.S.) of the equation:
\( \text{L.H.S.} = \vec{a} \times (\vec{b}+\vec{c}) + \vec{b} \times (\vec{c}+\vec{a}) + \vec{c} \times (\vec{a}+\vec{b}) \)
Using the distributive property of the cross product (\( \vec{x} \times (\vec{y}+\vec{z}) = \vec{x} \times \vec{y} + \vec{x} \times \vec{z} \)):
\( \implies \text{L.H.S.} = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) + (\vec{b} \times \vec{c} + \vec{b} \times \vec{a}) + (\vec{c} \times \vec{a} + \vec{c} \times \vec{b}) \)
Now, we use the property that \( \vec{x} \times \vec{y} = -(\vec{y} \times \vec{x}) \). This means \( \vec{a} \times \vec{c} = -(\vec{c} \times \vec{a}) \), \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \), and \( \vec{c} \times \vec{b} = -(\vec{b} \times \vec{c}) \).
Substitute these into the expression:
\( \implies \text{L.H.S.} = \vec{a} \times \vec{b} - (\vec{c} \times \vec{a}) + \vec{b} \times \vec{c} - (\vec{a} \times \vec{b}) + \vec{c} \times \vec{a} - (\vec{b} \times \vec{c}) \)
Rearrange the terms to group identical vectors:
\( \implies \text{L.H.S.} = (\vec{a} \times \vec{b} - \vec{a} \times \vec{b}) + (\vec{b} \times \vec{c} - \vec{b} \times \vec{c}) + (\vec{c} \times \vec{a} - \vec{c} \times \vec{a}) \)
Each pair cancels out:
\( \implies \text{L.H.S.} = \vec{0} + \vec{0} + \vec{0} \)
\( \implies \text{L.H.S.} = \vec{0} \)
Therefore, \( \vec{a} \times (\vec{b}+\vec{c}) + \vec{b} \times (\vec{c}+\vec{a}) + \vec{c} \times (\vec{a}+\vec{b}) = \vec{0} \). This proof highlights the anti-commutative and distributive properties of vector cross products.
In simple words: This proof shows that if you do a special kind of multiplication (cross product) for these three combinations of vectors and then add them up, the total answer will always be the zero vector. It uses rules about how cross products spread out and how their order changes their sign.

🎯 Exam Tip: Remember the two key properties for vector cross products: \( \vec{x} \times \vec{x} = \vec{0} \) and \( \vec{x} \times \vec{y} = -(\vec{y} \times \vec{x}) \). These are fundamental for simplifying such expressions.

Question 5. If \( \vec{a} \), \( \vec{b} \), \( \vec{c} \) are unit vectors, such that \( \vec{a} \cdot \vec{b} = 0 = \vec{a} \cdot \vec{c} \) and angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{6} \), then prove that \( \vec{a} = \pm 2 (\vec{b} \times \vec{c}) \).
Answer: Given that \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, this means their magnitudes are 1:
\( |\vec{a}| = 1, |\vec{b}| = 1, |\vec{c}| = 1 \)
We are given \( \vec{a} \cdot \vec{b} = 0 \). This means \( \vec{a} \) is perpendicular to \( \vec{b} \).
We are also given \( \vec{a} \cdot \vec{c} = 0 \). This means \( \vec{a} \) is perpendicular to \( \vec{c} \).
If a vector \( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \), then \( \vec{a} \) must be parallel to the cross product \( \vec{b} \times \vec{c} \).
So, we can write \( \vec{a} = \lambda (\vec{b} \times \vec{c}) \) for some scalar \( \lambda \). (Equation 1)
Now, let's find the magnitude of \( \vec{b} \times \vec{c} \). The angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{6} \).
\( |\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(\frac{\pi}{6}) \)
Since \( |\vec{b}| = 1 \) and \( |\vec{c}| = 1 \):
\( \implies |\vec{b} \times \vec{c}| = (1)(1)\sin(\frac{\pi}{6}) \)
\( \implies |\vec{b} \times \vec{c}| = \frac{1}{2} \)
From Equation 1, \( \vec{a} = \lambda (\vec{b} \times \vec{c}) \). Taking the magnitude of both sides:
\( |\vec{a}| = |\lambda (\vec{b} \times \vec{c})| \)
\( \implies |\vec{a}| = |\lambda| |\vec{b} \times \vec{c}| \)
We know \( |\vec{a}| = 1 \) and \( |\vec{b} \times \vec{c}| = \frac{1}{2} \):
\( \implies 1 = |\lambda| \left(\frac{1}{2}\right) \)
\( \implies |\lambda| = 2 \)
Therefore, \( \lambda = \pm 2 \).
Substituting the value of \( \lambda \) back into Equation 1:
\( \vec{a} = \pm 2 (\vec{b} \times \vec{c}) \)
This demonstrates how the geometric properties of vectors (perpendicularity, unit length, angle) directly translate into an algebraic relationship between them.
In simple words: Since vector 'a' is perpendicular to both 'b' and 'c' (meaning their dot products are zero), it must point in the same direction as the cross product of 'b' and 'c'. Because all are "unit vectors" (length 1) and the angle between 'b' and 'c' is known, we can figure out the exact number (which is ±2) that connects vector 'a' to the cross product of 'b' and 'c'.

🎯 Exam Tip: When a vector is perpendicular to two other non-parallel vectors, it must be parallel to their cross product. This is a crucial concept in vector algebra.

Question 6. Find the value of \( |\vec{a} \times \vec{b}| \), if \( |\vec{a}| = 10, |\vec{b}|= 2 \) and \( \vec{a} \cdot \vec{b} = 12 \).
Answer: We are given:
\( |\vec{a}| = 10 \)
\( |\vec{b}| = 2 \)
\( \vec{a} \cdot \vec{b} = 12 \)
We know the identity relating the dot product, cross product, and magnitudes of two vectors (Lagrange's Identity):
\( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \)
Substitute the given values into this identity:
\( |\vec{a} \times \vec{b}|^2 + (12)^2 = (10)^2 (2)^2 \)
\( \implies |\vec{a} \times \vec{b}|^2 + 144 = (100)(4) \)
\( \implies |\vec{a} \times \vec{b}|^2 + 144 = 400 \)
Now, subtract 144 from both sides to solve for \( |\vec{a} \times \vec{b}|^2 \):
\( \implies |\vec{a} \times \vec{b}|^2 = 400 - 144 \)
\( \implies |\vec{a} \times \vec{b}|^2 = 256 \)
Take the square root of both sides to find \( |\vec{a} \times \vec{b}| \):
\( \implies |\vec{a} \times \vec{b}| = \sqrt{256} \)
\( \implies |\vec{a} \times \vec{b}| = 16 \)
The magnitude of the cross product \( |\vec{a} \times \vec{b}| \) is 16. This formula provides an alternative way to calculate the cross product's magnitude without needing the angle between the vectors directly.
In simple words: We are given the lengths of two vectors and their "dot product." We use a special rule that links the dot product, cross product, and lengths of vectors. By putting in the numbers we know, we can find the length of their "cross product," which turns out to be 16.

🎯 Exam Tip: Always remember Lagrange's Identity for vectors \( |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \), as it simplifies many vector problems where angle information might be missing.

Question 7. Find vector perpendicular to the vectors \( 4\hat{i}-\hat{j}+3\hat{k} \) and \( 2\hat{i}+\hat{j}-2\hat{k} \).
Answer: Let \( \vec{a} = 4\hat{i}-\hat{j}+3\hat{k} \) and \( \vec{b} = 2\hat{i}+\hat{j}-2\hat{k} \).
A vector perpendicular to two given vectors is their cross product.
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ 2 & 1 & -2 \end{vmatrix} \)
Now, we expand the determinant:
\( \implies \vec{a} \times \vec{b} = \hat{i}((-1) \times (-2) - 3 \times 1) - \hat{j}(4 \times (-2) - 3 \times 2) + \hat{k}(4 \times 1 - (-1) \times 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(2 - 3) - \hat{j}(-8 - 6) + \hat{k}(4 + 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(-1) - \hat{j}(-14) + \hat{k}(6) \)
\( \implies \vec{a} \times \vec{b} = -\hat{i} + 14\hat{j} + 6\hat{k} \)
This vector, \( -\hat{i} + 14\hat{j} + 6\hat{k} \), is perpendicular to both \( \vec{a} \) and \( \vec{b} \). This is a direct application of the cross product's definition.
In simple words: To find a vector that is perpendicular (at a right angle) to two other vectors, we simply calculate their "cross product." This mathematical operation gives us a new vector that points in the direction perpendicular to both original vectors.

🎯 Exam Tip: The cross product \( \vec{a} \times \vec{b} \) always results in a vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \). This property is fundamental in physics and geometry.

Question 8. Show that: \( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \), also, explain geometrically.
Answer: **Algebraic Proof:**
We start with the Left Hand Side (L.H.S.):
\( \text{L.H.S.} = (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) \)
Using the distributive property of the cross product:
\( \implies \text{L.H.S.} = \vec{a} \times (\vec{a}+\vec{b}) - \vec{b} \times (\vec{a}+\vec{b}) \)
\( \implies \text{L.H.S.} = (\vec{a} \times \vec{a} + \vec{a} \times \vec{b}) - (\vec{b} \times \vec{a} + \vec{b} \times \vec{b}) \)
We know that the cross product of a vector with itself is the zero vector (\( \vec{x} \times \vec{x} = \vec{0} \)) and \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \):
\( \implies \text{L.H.S.} = (\vec{0} + \vec{a} \times \vec{b}) - (-(\vec{a} \times \vec{b}) + \vec{0}) \)
\( \implies \text{L.H.S.} = \vec{a} \times \vec{b} - (-\vec{a} \times \vec{b}) \)
\( \implies \text{L.H.S.} = \vec{a} \times \vec{b} + \vec{a} \times \vec{b} \)
\( \implies \text{L.H.S.} = 2(\vec{a} \times \vec{b}) \)
Thus, \( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \).
**Geometrical Interpretation:**
Consider a parallelogram ABCD with adjacent sides represented by vectors \( \vec{AB} = \vec{a} \) and \( \vec{AD} = \vec{b} \).
The diagonals of the parallelogram are:
\( \vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b} \) (using triangle law of vector addition)
\( \vec{DB} = \vec{AB} - \vec{AD} = \vec{a} - \vec{b} \) (using triangle law, \( \vec{AD} + \vec{DB} = \vec{AB} \implies \vec{DB} = \vec{AB} - \vec{AD} \))
The vector area of the parallelogram formed by diagonals \( \vec{DB} \) and \( \vec{AC} \) is given by \( \frac{1}{2} (\vec{DB} \times \vec{AC}) \).
So, \( \text{Vector area} = \frac{1}{2} ((\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})) \)
From our algebraic proof, we know that \( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \).
Substitute this into the area formula:
\( \text{Vector area} = \frac{1}{2} (2(\vec{a} \times \vec{b})) \)
\( \text{Vector area} = \vec{a} \times \vec{b} \)
The vector area of the parallelogram formed by adjacent sides \( \vec{a} \) and \( \vec{b} \) is also \( \vec{a} \times \vec{b} \).
Therefore, the result \( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \) geometrically means that the vector area of a parallelogram whose adjacent sides are the diagonals of the original parallelogram, is twice the vector area of the original parallelogram.
In simple words: First, we prove the equation using simple vector rules like how cross products spread out and how a vector crossed with itself is zero. Geometrically, this equation means that if you make a new parallelogram using the diagonals of an original one, the area of this new parallelogram is twice the area of the first one.

🎯 Exam Tip: When explaining geometrically, drawing a clear diagram with labeled vectors and diagonals can significantly help in visualizing the vector relationships and area interpretation.

Question 9. For any vector \( \vec{a} \), prove that \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \).
Answer: Let \( \vec{a} \) be any vector. We can write it in component form as:
\( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \)
The magnitude of \( \vec{a} \) is \( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \), so \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \).
Now, let's calculate each term on the Left Hand Side (L.H.S.):
1. \( \vec{a} \times \hat{i} \):
\( \vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i} \)
\( \implies \vec{a} \times \hat{i} = a_1(\hat{i} \times \hat{i}) + a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) \)
We know \( \hat{i} \times \hat{i} = \vec{0} \), \( \hat{j} \times \hat{i} = -\hat{k} \), and \( \hat{k} \times \hat{i} = \hat{j} \):
\( \implies \vec{a} \times \hat{i} = a_1(\vec{0}) + a_2(-\hat{k}) + a_3(\hat{j}) \)
\( \implies \vec{a} \times \hat{i} = a_3\hat{j} - a_2\hat{k} \)
The magnitude squared is:
\( |\vec{a} \times \hat{i}|^2 = (a_3)^2 + (-a_2)^2 = a_3^2 + a_2^2 \)
2. \( \vec{a} \times \hat{j} \):
\( \vec{a} \times \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{j} \)
\( \implies \vec{a} \times \hat{j} = a_1(\hat{i} \times \hat{j}) + a_2(\hat{j} \times \hat{j}) + a_3(\hat{k} \times \hat{j}) \)
We know \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{j} = \vec{0} \), and \( \hat{k} \times \hat{j} = -\hat{i} \):
\( \implies \vec{a} \times \hat{j} = a_1(\hat{k}) + a_2(\vec{0}) + a_3(-\hat{i}) \)
\( \implies \vec{a} \times \hat{j} = a_1\hat{k} - a_3\hat{i} \)
The magnitude squared is:
\( |\vec{a} \times \hat{j}|^2 = (-a_3)^2 + (a_1)^2 = a_3^2 + a_1^2 \)
3. \( \vec{a} \times \hat{k} \):
\( \vec{a} \times \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{k} \)
\( \implies \vec{a} \times \hat{k} = a_1(\hat{i} \times \hat{k}) + a_2(\hat{j} \times \hat{k}) + a_3(\hat{k} \times \hat{k}) \)
We know \( \hat{i} \times \hat{k} = -\hat{j} \), \( \hat{j} \times \hat{k} = \hat{i} \), and \( \hat{k} \times \hat{k} = \vec{0} \):
\( \implies \vec{a} \times \hat{k} = a_1(-\hat{j}) + a_2(\hat{i}) + a_3(\vec{0}) \)
\( \implies \vec{a} \times \hat{k} = a_2\hat{i} - a_1\hat{j} \)
The magnitude squared is:
\( |\vec{a} \times \hat{k}|^2 = (a_2)^2 + (-a_1)^2 = a_2^2 + a_1^2 \)
Now, add the three squared magnitudes:
\( \text{L.H.S.} = |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 \)
\( \implies \text{L.H.S.} = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) \)
\( \implies \text{L.H.S.} = 2a_1^2 + 2a_2^2 + 2a_3^2 \)
Factor out 2:
\( \implies \text{L.H.S.} = 2(a_1^2 + a_2^2 + a_3^2) \)
Since \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \):
\( \implies \text{L.H.S.} = 2|\vec{a}|^2 \)
This proves the identity. This identity is a good illustration of how vector components interact with standard basis vectors in cross products.
In simple words: This proof shows a special rule for any vector. If you take a vector and find its "cross product" with each of the basic direction vectors (\( \hat{i}, \hat{j}, \hat{k} \)), then square the length of these results and add them up, the total will always be two times the square of the original vector's length.

🎯 Exam Tip: Remember the cyclic properties of cross products for unit vectors: \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \), and reversing the order changes the sign (e.g., \( \hat{j} \times \hat{i} = -\hat{k} \)). Also, \( \hat{i} \times \hat{i} = \vec{0} \).

Question 10. If two adjacent sides of a triangle are represented by vectors \( \hat{i}+2\hat{j}+2\hat{k} \) and \( 3\hat{i}-2\hat{j}+\hat{k} \), then find the area of triangle.
Answer: Let the two adjacent sides of the triangle be represented by vectors \( \vec{a} = \hat{i}+2\hat{j}+2\hat{k} \) and \( \vec{b} = 3\hat{i}-2\hat{j}+\hat{k} \).
The area of a triangle with adjacent sides \( \vec{a} \) and \( \vec{b} \) is given by half the magnitude of their cross product:
\( \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| \)
First, calculate the cross product \( \vec{a} \times \vec{b} \):
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & -2 & 1 \end{vmatrix} \)
\( \implies \vec{a} \times \vec{b} = \hat{i}((2) \times (1) - (2) \times (-2)) - \hat{j}((1) \times (1) - (2) \times (3)) + \hat{k}((1) \times (-2) - (2) \times (3)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(2 - (-4)) - \hat{j}(1 - 6) + \hat{k}(-2 - 6) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(2 + 4) - \hat{j}(-5) + \hat{k}(-8) \)
\( \implies \vec{a} \times \vec{b} = 6\hat{i} + 5\hat{j} - 8\hat{k} \)
Next, find the magnitude of this cross product:
\( |\vec{a} \times \vec{b}| = \sqrt{6^2 + 5^2 + (-8)^2} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{36 + 25 + 64} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{125} \)
\( \implies |\vec{a} \times \vec{b}| = \sqrt{25 \times 5} = 5\sqrt{5} \)
Finally, calculate the area of the triangle:
\( \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| \)
\( \implies \text{Area} = \frac{1}{2} (5\sqrt{5}) \)
\( \implies \text{Area} = \frac{5\sqrt{5}}{2} \) square units.
This method efficiently calculates the area of a triangle given two of its sides as vectors, without needing to know any angles.
In simple words: To find the area of a triangle when you know two of its sides as vectors, you first do a special multiplication called the "cross product" of these two vectors. Then, you find the length of the new vector you get and divide that length by two. This gives you the triangle's area.

🎯 Exam Tip: Always remember that the area of a triangle formed by two vectors \( \vec{a} \) and \( \vec{b} \) as adjacent sides is \( \frac{1}{2} |\vec{a} \times \vec{b}| \), while the area of a parallelogram is \( |\vec{a} \times \vec{b}| \).

Rajasthan Board RBSE Class 12 Maths Chapter 13 Vector Ex 13.3

Question 1. Find vector product of vectors \( 3\hat{i}+\hat{j}-\hat{k} \) and \( 2\hat{i}+3\hat{j}+\hat{k} \).
Answer: Let the given vectors be \( \vec{a} = 3\hat{i}+\hat{j}-\hat{k} \) and \( \vec{b} = 2\hat{i}+3\hat{j}+\hat{k} \).
To find the vector product \( \vec{a} \times \vec{b} \), we calculate the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of \( \vec{a} \) and \( \vec{b} \). This calculation gives us a new vector that is perpendicular to both original vectors.
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -1 \\ 2 & 3 & 1 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = \hat{i}(1 \times 1 - (-1) \times 3) - \hat{j}(3 \times 1 - (-1) \times 2) + \hat{k}(3 \times 3 - 1 \times 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(1+3) - \hat{j}(3+2) + \hat{k}(9-2) \)
\( \implies \vec{a} \times \vec{b} = 4\hat{i} - 5\hat{j} + 7\hat{k} \)
Therefore, the cross product of the vectors \( 3\hat{i}+\hat{j}-\hat{k} \) and \( 2\hat{i}+3\hat{j}+\hat{k} \) is \( 4\hat{i}-5\hat{j}+7\hat{k} \).
In simple words: To find the vector product, we arrange the components in a special grid and calculate a determinant. This gives us a new vector, which is perpendicular to the first two.

🎯 Exam Tip: Remember the sign pattern \( (+, -, +) \) when expanding the determinant for cross products to avoid common calculation errors.

Question 2. Find perpendicular unit vector of vectors \( \hat{i}-2\hat{j}+\hat{k} \) and \( 2\hat{i}+\hat{j}-3\hat{k} \).
Answer: Let the given vectors be \( \vec{a} = \hat{i}-2\hat{j}+\hat{k} \) and \( \vec{b} = 2\hat{i}+\hat{j}-3\hat{k} \).
First, we find the cross product of the two vectors, which gives a vector perpendicular to both. A unit vector in this direction is found by dividing this vector by its magnitude.
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 2 & 1 & -3 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = \hat{i}((-2) \times (-3) - 1 \times 1) - \hat{j}(1 \times (-3) - 1 \times 2) + \hat{k}(1 \times 1 - (-2) \times 2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(6-1) - \hat{j}(-3-2) + \hat{k}(1+4) \)
\( \implies \vec{a} \times \vec{b} = 5\hat{i} + 5\hat{j} + 5\hat{k} \)
Next, we find the magnitude of this cross product:
\( | \vec{a} \times \vec{b} | = \sqrt{5^2 + 5^2 + 5^2} \)
\( \implies | \vec{a} \times \vec{b} | = \sqrt{25+25+25} \)
\( \implies | \vec{a} \times \vec{b} | = \sqrt{75} \)
\( \implies | \vec{a} \times \vec{b} | = \sqrt{25 \times 3} = 5\sqrt{3} \)
Finally, the perpendicular unit vector \( \hat{n} \) is the cross product divided by its magnitude:
\( \hat{n} = \frac{\vec{a} \times \vec{b}}{| \vec{a} \times \vec{b} |} = \frac{5\hat{i} + 5\hat{j} + 5\hat{k}}{5\sqrt{3}} \)
\( \implies \hat{n} = \frac{5(\hat{i} + \hat{j} + \hat{k})}{5\sqrt{3}} \)
\( \implies \hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \)
So, the perpendicular unit vector to \( \hat{i}-2\hat{j}+\hat{k} \) and \( 2\hat{i}+\hat{j}-3\hat{k} \) is \( \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} \).
In simple words: First, we multiply the two vectors in a special way to get a new vector that stands straight up from both. Then, we find its length and divide the vector by this length to get a "unit" vector, which is a vector of length one pointing in the same direction.

🎯 Exam Tip: A unit vector always has a magnitude of 1. If your final vector does not have a magnitude of 1, double-check your calculations.

Question 3. For vectors \( \vec{a} \) and \( \vec{b} \), prove that \( | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 - (\vec{a} \cdot \vec{b})^2 \).
Answer: Let \( \theta \) be the angle between vectors \( \vec{a} \) and \( \vec{b} \).
We know the definition of the cross product magnitude and dot product:
\( | \vec{a} \times \vec{b} | = | \vec{a} | | \vec{b} | \sin\theta \hat{n} \)
Where \( \hat{n} \) is a unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \).
Therefore, \( | \vec{a} \times \vec{b} |^2 = (| \vec{a} | | \vec{b} | \sin\theta \hat{n}) \cdot (| \vec{a} | | \vec{b} | \sin\theta \hat{n}) \)
\( \implies | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 \sin^2\theta (\hat{n} \cdot \hat{n}) \)
Since \( \hat{n} \) is a unit vector, \( \hat{n} \cdot \hat{n} = |\hat{n}|^2 = 1 \).
\( \implies | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 \sin^2\theta \)
We also know the trigonometric identity \( \sin^2\theta = 1 - \cos^2\theta \).
\( \implies | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 (1 - \cos^2\theta) \)
\( \implies | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 - | \vec{a} |^2 | \vec{b} |^2 \cos^2\theta \)
From the definition of the dot product, \( \vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos\theta \).
So, \( (\vec{a} \cdot \vec{b})^2 = (| \vec{a} | | \vec{b} | \cos\theta)^2 = | \vec{a} |^2 | \vec{b} |^2 \cos^2\theta \).
Substituting this into the equation for \( | \vec{a} \times \vec{b} |^2 \):
\( | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 - (\vec{a} \cdot \vec{b})^2 \)
This identity is also known as Lagrange's Identity for vectors. It connects the dot product and cross product magnitudes.
\[ \text{Alternative proof using determinant identity for } (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) \] \[ (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} \end{vmatrix} \] \[ \implies | \vec{a} \times \vec{b} |^2 = (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b}) - (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{a}) \] \[ \implies | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 - (\vec{a} \cdot \vec{b})^2 \] Thus, the identity is proven.
In simple words: This rule shows a connection between how much two vectors spread apart (cross product) and how much they point in the same direction (dot product). It says that the squared "spread" can be found by subtracting the squared "alignment" from the product of their squared lengths.

🎯 Exam Tip: This identity is fundamental in vector algebra. Remember that \( \sin^2\theta + \cos^2\theta = 1 \) is key to deriving it.

Question 4. Prove that \( \vec{a} \times (\vec{b}+\vec{c}) + \vec{b} \times (\vec{c}+\vec{a}) + \vec{c} \times (\vec{a} + \vec{b}) = \vec{0} \).
Answer: We need to prove that the left-hand side (L.H.S.) equals the right-hand side (R.H.S.), which is the zero vector \( \vec{0} \).
Let's expand the terms on the L.H.S. using the distributive property of the cross product:
\( \text{L.H.S.} = \vec{a} \times (\vec{b}+\vec{c}) + \vec{b} \times (\vec{c}+\vec{a}) + \vec{c} \times (\vec{a} + \vec{b}) \)
\( \implies \text{L.H.S.} = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) + (\vec{b} \times \vec{c} + \vec{b} \times \vec{a}) + (\vec{c} \times \vec{a} + \vec{c} \times \vec{b}) \)
Now, we use the property that \( \vec{x} \times \vec{y} = - (\vec{y} \times \vec{x}) \). So, \( \vec{a} \times \vec{c} = - (\vec{c} \times \vec{a}) \), \( \vec{b} \times \vec{a} = - (\vec{a} \times \vec{b}) \), and \( \vec{c} \times \vec{b} = - (\vec{b} \times \vec{c}) \).
Substitute these into the expanded expression:
\( \text{L.H.S.} = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} - (\vec{a} \times \vec{b}) + \vec{c} \times \vec{a} - (\vec{b} \times \vec{c}) \)
Rearranging the terms to group common cross products:
\( \text{L.H.S.} = (\vec{a} \times \vec{b} - \vec{a} \times \vec{b}) + (\vec{a} \times \vec{c} + \vec{c} \times \vec{a}) + (\vec{b} \times \vec{c} - \vec{b} \times \vec{c}) \)
As we know, \( \vec{a} \times \vec{c} = - (\vec{c} \times \vec{a}) \), so \( \vec{a} \times \vec{c} + \vec{c} \times \vec{a} = \vec{0} \). Also, a vector cross-product with itself is zero, so \( \vec{x} \times \vec{x} = \vec{0} \).
\( \text{L.H.S.} = \vec{0} + \vec{0} + \vec{0} \)
\( \implies \text{L.H.S.} = \vec{0} \)
Since L.H.S. = R.H.S., the identity is proven. This property demonstrates how cross products distribute and cancel out in a cyclic sum.
In simple words: When you take the cross product of one vector with the sum of two others, and do this for all possible pairs in a cycle, the results all cancel each other out, adding up to nothing.

🎯 Exam Tip: Remember the distributive property \( \vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C} \) and the anti-commutative property \( \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}) \) when solving such proofs.

Question 5. If \( \vec{a} \), \( \vec{b} \), \( \vec{c} \) are unit vectors, such that \( \vec{a} \cdot \vec{b} = 0 = \vec{a} \cdot \vec{c} \) and angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{6} \), then prove that \( \vec{a} = \pm 2 (\vec{b} \times \vec{c}) \).
Answer: We are given that \( \vec{a} \), \( \vec{b} \), \( \vec{c} \) are unit vectors, which means their magnitudes are 1: \( | \vec{a} | = | \vec{b} | = | \vec{c} | = 1 \).
We are also given that \( \vec{a} \cdot \vec{b} = 0 \) and \( \vec{a} \cdot \vec{c} = 0 \).
This implies that vector \( \vec{a} \) is perpendicular to both vector \( \vec{b} \) and vector \( \vec{c} \).
We know that the cross product \( \vec{b} \times \vec{c} \) results in a vector that is perpendicular to both \( \vec{b} \) and \( \vec{c} \).
Therefore, vector \( \vec{a} \) must be parallel to the vector \( \vec{b} \times \vec{c} \).
If two vectors are parallel, one can be expressed as a scalar multiple of the other.
Let \( \vec{a} = \lambda (\vec{b} \times \vec{c}) \), where \( \lambda \) is some scalar.
Now, let's find the magnitude of \( \vec{b} \times \vec{c} \). The angle between \( \vec{b} \) and \( \vec{c} \) is given as \( \frac{\pi}{6} \).
\( | \vec{b} \times \vec{c} | = | \vec{b} | | \vec{c} | \sin(\frac{\pi}{6}) \)
Since \( | \vec{b} | = 1 \) and \( | \vec{c} | = 1 \), and \( \sin(\frac{\pi}{6}) = \frac{1}{2} \):
\( | \vec{b} \times \vec{c} | = (1)(1)(\frac{1}{2}) = \frac{1}{2} \).
Now we take the magnitude of our equation \( \vec{a} = \lambda (\vec{b} \times \vec{c}) \):
\( | \vec{a} | = | \lambda | | \vec{b} \times \vec{c} | \)
We know \( | \vec{a} | = 1 \) and \( | \vec{b} \times \vec{c} | = \frac{1}{2} \).
\( 1 = | \lambda | (\frac{1}{2}) \)
\( \implies | \lambda | = 2 \)
\( \implies \lambda = \pm 2 \)
Substitute the value of \( \lambda \) back into the equation for \( \vec{a} \):
\( \vec{a} = \pm 2 (\vec{b} \times \vec{c}) \)
Thus, the given statement is proven. This shows the strong geometric relationship between mutually orthogonal unit vectors and their cross products.
In simple words: We are given that vector 'a' is a unit vector and is at right angles to both 'b' and 'c'. Also, 'b' and 'c' are unit vectors with a small angle between them. Because 'a' is perpendicular to both 'b' and 'c', it must be pointing in the same direction as 'b' cross 'c'. By comparing their lengths, we find that 'a' is exactly twice the length of 'b' cross 'c', and it can point in either of the two opposite directions.

🎯 Exam Tip: When vectors are perpendicular, their dot product is zero. When a vector is parallel to another, it is a scalar multiple of that vector. These are key concepts for such proofs.

Question 6. Find the value of \( | \vec{a} \times \vec{b} | \), if \( | \vec{a} | = 10, | \vec{b} | = 2 \) and \( \vec{a} \cdot \vec{b} = 12 \).
Answer: We are given the magnitudes of vectors \( \vec{a} \) and \( \vec{b} \), and their dot product:
\( | \vec{a} | = 10 \)
\( | \vec{b} | = 2 \)
\( \vec{a} \cdot \vec{b} = 12 \)
We need to find the magnitude of the cross product, \( | \vec{a} \times \vec{b} | \).
We can use the identity we proved in Question 3:
\( | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 - (\vec{a} \cdot \vec{b})^2 \)
Substitute the given values into this identity:
\( | \vec{a} \times \vec{b} |^2 = (10)^2 (2)^2 - (12)^2 \)
\( \implies | \vec{a} \times \vec{b} |^2 = (100)(4) - 144 \)
\( \implies | \vec{a} \times \vec{b} |^2 = 400 - 144 \)
\( \implies | \vec{a} \times \vec{b} |^2 = 256 \)
Now, take the square root of both sides to find \( | \vec{a} \times \vec{b} | \):
\( | \vec{a} \times \vec{b} | = \sqrt{256} \)
\( \implies | \vec{a} \times \vec{b} | = 16 \)
Alternatively, we could first find the angle \( \theta \) between the vectors using the dot product formula:
\( \vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos\theta \)
\( 12 = (10)(2) \cos\theta \)
\( 12 = 20 \cos\theta \)
\( \implies \cos\theta = \frac{12}{20} = \frac{3}{5} \)
Now, use the identity \( \sin\theta = \sqrt{1 - \cos^2\theta} \) (since \( \theta \) is typically taken as \( 0 \le \theta \le \pi \), \( \sin\theta \ge 0 \)).
\( \sin\theta = \sqrt{1 - (\frac{3}{5})^2} \)
\( \implies \sin\theta = \sqrt{1 - \frac{9}{25}} \)
\( \implies \sin\theta = \sqrt{\frac{25-9}{25}} = \sqrt{\frac{16}{25}} \)
\( \implies \sin\theta = \frac{4}{5} \)
Finally, use the cross product magnitude formula:
\( | \vec{a} \times \vec{b} | = | \vec{a} | | \vec{b} | \sin\theta \)
\( \implies | \vec{a} \times \vec{b} | = (10)(2)(\frac{4}{5}) \)
\( \implies | \vec{a} \times \vec{b} | = 20 \times \frac{4}{5} \)
\( \implies | \vec{a} \times \vec{b} | = 4 \times 4 = 16 \)
Both methods yield the same result. The first method using the identity is generally quicker.
In simple words: We are given the lengths of two vectors and how much they point in the same direction (their dot product). We use a special formula that links these values to how much the vectors spread apart (their cross product). After plugging in the numbers, we find the "spread" value.

🎯 Exam Tip: Always remember the identity \( | \vec{a} \times \vec{b} |^2 = | \vec{a} |^2 | \vec{b} |^2 - (\vec{a} \cdot \vec{b})^2 \) as it provides a direct way to find the cross product magnitude when the dot product is known.

Question 7. Find vector perpendicular to the vectors \( 4\hat{i}-\hat{j}+3\hat{k} \) and \( -2\hat{i}+\hat{j}-2\hat{k} \).
Answer: Let the given vectors be \( \vec{a} = 4\hat{i}-\hat{j}+3\hat{k} \) and \( \vec{b} = -2\hat{i}+\hat{j}-2\hat{k} \).
A vector perpendicular to two given vectors is found by their cross product. The cross product \( \vec{a} \times \vec{b} \) yields a vector that is orthogonal (perpendicular) to both \( \vec{a} \) and \( \vec{b} \).
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -1 & 3 \\ -2 & 1 & -2 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = \hat{i}((-1) \times (-2) - 3 \times 1) - \hat{j}(4 \times (-2) - 3 \times (-2)) + \hat{k}(4 \times 1 - (-1) \times (-2)) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(2-3) - \hat{j}(-8+6) + \hat{k}(4-2) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(-1) - \hat{j}(-2) + \hat{k}(2) \)
\( \implies \vec{a} \times \vec{b} = -\hat{i} + 2\hat{j} + 2\hat{k} \)
So, a vector perpendicular to the given vectors is \( -\hat{i} + 2\hat{j} + 2\hat{k} \). We could also find a unit vector in this direction by dividing by its magnitude, but the question only asks for "a vector".
In simple words: To find a vector that is at right angles to two other vectors, we perform a special multiplication called the cross product. This operation gives us the desired perpendicular vector.

🎯 Exam Tip: Remember that the cross product of two vectors is always perpendicular to the plane containing those two vectors. Pay careful attention to the signs in the determinant expansion.

Question 8. Show that: \( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \) also, explain geometrically.
Answer: To prove the identity, we expand the left-hand side (L.H.S.) using the distributive property of the cross product:
\( \text{L.H.S.} = (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) \)
\( \implies \text{L.H.S.} = \vec{a} \times (\vec{a}+\vec{b}) - \vec{b} \times (\vec{a}+\vec{b}) \)
\( \implies \text{L.H.S.} = (\vec{a} \times \vec{a} + \vec{a} \times \vec{b}) - (\vec{b} \times \vec{a} + \vec{b} \times \vec{b}) \)
We know that the cross product of a vector with itself is the zero vector, i.e., \( \vec{x} \times \vec{x} = \vec{0} \). So, \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{b} = \vec{0} \).
Also, we use the anti-commutative property of the cross product: \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \).
Substitute these properties into the expression:
\( \text{L.H.S.} = (\vec{0} + \vec{a} \times \vec{b}) - (-(\vec{a} \times \vec{b}) + \vec{0}) \)
\( \implies \text{L.H.S.} = \vec{a} \times \vec{b} + \vec{a} \times \vec{b} \)
\( \implies \text{L.H.S.} = 2(\vec{a} \times \vec{b}) \)
This matches the right-hand side (R.H.S.). Hence, the identity is proven.

Geometrical Interpretation:
Let ABCD be a parallelogram with adjacent sides \( \vec{AB} = \vec{a} \) and \( \vec{AD} = \vec{b} \).
The diagonal \( \vec{AC} \) (vector sum) is \( \vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b} \).
The diagonal \( \vec{DB} \) (vector difference) is \( \vec{DB} = \vec{AB} - \vec{AD} = \vec{a} - \vec{b} \).
(Note: Some texts define the other diagonal as \( \vec{BD} = \vec{AD} - \vec{AB} = \vec{b} - \vec{a} \), which would mean \( -\vec{DB} \). The problem uses \( (\vec{a}-\vec{b}) \), so we align with \( \vec{DB} \)).

The vector area of the parallelogram formed by diagonals \( \vec{DB} \) and \( \vec{AC} \) is given by \( \frac{1}{2} (\vec{DB} \times \vec{AC}) \).
\[ \vec{DB} \times \vec{AC} = (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) \] We just proved that \( (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b}) \).
So, \( \vec{DB} \times \vec{AC} = 2(\vec{a} \times \vec{b}) \).
The vector area of the parallelogram formed by the adjacent sides \( \vec{a} \) and \( \vec{b} \) is \( \vec{a} \times \vec{b} \).
Therefore, \( \vec{DB} \times \vec{AC} = 2 \times (\text{Vector area of parallelogram ABCD}) \).
This means that the vector cross product of the diagonals of a parallelogram is twice the vector area of the parallelogram itself. The area of a parallelogram formed by two vectors is a crucial application of the cross product.
In simple words: This means if you take two vectors that form the sides of a parallelogram, and then look at the parallelogram's diagonals, the cross product of these diagonals is equal to twice the area of the parallelogram itself. It shows how the diagonals relate to the total area.

🎯 Exam Tip: Remember that \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{a} \times \vec{b} = -\vec{b} \times \vec{a} \) are crucial properties for simplifying vector cross product expressions.

Question 9. For any vector \( \vec{a} \), prove that \( |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = 2|\vec{a}|^2 \).
Answer: Let vector \( \vec{a} \) be represented by its components: \( \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \).
The magnitude squared of \( \vec{a} \) is \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \).
Now, let's calculate each term on the left-hand side (L.H.S.):
First term: \( \vec{a} \times \hat{i} \)
\[ \vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i} \]
\( \implies \vec{a} \times \hat{i} = a_1(\hat{i} \times \hat{i}) + a_2(\hat{j} \times \hat{i}) + a_3(\hat{k} \times \hat{i}) \)
Using the cross product rules (\( \hat{i} \times \hat{i} = \vec{0} \), \( \hat{j} \times \hat{i} = -\hat{k} \), \( \hat{k} \times \hat{i} = \hat{j} \)):
\( \implies \vec{a} \times \hat{i} = a_1(\vec{0}) + a_2(-\hat{k}) + a_3(\hat{j}) = a_3\hat{j} - a_2\hat{k} \)
So, \( |\vec{a} \times \hat{i}|^2 = (a_3)^2 + (-a_2)^2 = a_3^2 + a_2^2 \).

Second term: \( \vec{a} \times \hat{j} \)
\[ \vec{a} \times \hat{j} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{j} \]
\( \implies \vec{a} \times \hat{j} = a_1(\hat{i} \times \hat{j}) + a_2(\hat{j} \times \hat{j}) + a_3(\hat{k} \times \hat{j}) \)
Using the cross product rules (\( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{j} = \vec{0} \), \( \hat{k} \times \hat{j} = -\hat{i} \)):
\( \implies \vec{a} \times \hat{j} = a_1(\hat{k}) + a_2(\vec{0}) + a_3(-\hat{i}) = -a_3\hat{i} + a_1\hat{k} \)
So, \( |\vec{a} \times \hat{j}|^2 = (-a_3)^2 + (a_1)^2 = a_3^2 + a_1^2 \).

Third term: \( \vec{a} \times \hat{k} \)
\[ \vec{a} \times \hat{k} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{k} \]
\( \implies \vec{a} \times \hat{k} = a_1(\hat{i} \times \hat{k}) + a_2(\hat{j} \times \hat{k}) + a_3(\hat{k} \times \hat{k}) \)
Using the cross product rules (\( \hat{i} \times \hat{k} = -\hat{j} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{k} = \vec{0} \)):
\( \implies \vec{a} \times \hat{k} = a_1(-\hat{j}) + a_2(\hat{i}) + a_3(\vec{0}) = a_2\hat{i} - a_1\hat{j} \)
So, \( |\vec{a} \times \hat{k}|^2 = (a_2)^2 + (-a_1)^2 = a_2^2 + a_1^2 \).

Now, sum these three squared magnitudes:
\( \text{L.H.S.} = |\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 \)
\( \implies \text{L.H.S.} = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) \)
\( \implies \text{L.H.S.} = 2a_1^2 + 2a_2^2 + 2a_3^2 \)
\( \implies \text{L.H.S.} = 2(a_1^2 + a_2^2 + a_3^2) \)
Since \( |\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2 \):
\( \implies \text{L.H.S.} = 2|\vec{a}|^2 \)
This equals the right-hand side (R.H.S.). The identity is proven. This identity shows a relationship between the magnitude of a vector and its cross products with the standard basis vectors.
In simple words: This proof shows that if you take any vector, find its cross product with the three main direction vectors (\( \hat{i}, \hat{j}, \hat{k} \)), square the lengths of these three results, and then add them up, the total will be exactly twice the square of the original vector's length.

🎯 Exam Tip: Systematically calculate each cross product and its magnitude. Remember the cyclic properties of unit vector cross products (\( \hat{i} \times \hat{j} = \hat{k} \), etc.) and that \( \hat{i} \times \hat{i} = \vec{0} \).

Question 10. If two adjacent sides of a triangle are represented by vectors \( \hat{i}+2\hat{j}+2\hat{k} \) and \( 3\hat{i}-2\hat{j}+\hat{k} \), then find the area of triangle.
Answer: Let the two adjacent sides of the triangle be represented by vectors \( \vec{a} = \hat{i}+2\hat{j}+2\hat{k} \) and \( \vec{b} = 3\hat{i}-2\hat{j}+\hat{k} \).
The area of a triangle whose adjacent sides are given by vectors \( \vec{a} \) and \( \vec{b} \) is half the magnitude of their cross product. This is because the magnitude of the cross product \( | \vec{a} \times \vec{b} | \) represents the area of the parallelogram formed by \( \vec{a} \) and \( \vec{b} \), and a triangle is half of a parallelogram.
First, find the cross product \( \vec{a} \times \vec{b} \):
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & -2 & 1 \end{vmatrix} \]
\( \implies \vec{a} \times \vec{b} = \hat{i}(2 \times 1 - 2 \times (-2)) - \hat{j}(1 \times 1 - 2 \times 3) + \hat{k}(1 \times (-2) - 2 \times 3) \)
\( \implies \vec{a} \times \vec{b} = \hat{i}(2+4) - \hat{j}(1-6) + \hat{k}(-2-6) \)
\( \implies \vec{a} \times \vec{b} = 6\hat{i} - (-5\hat{j}) + (-8\hat{k}) \)
\( \implies \vec{a} \times \vec{b} = 6\hat{i} + 5\hat{j} - 8\hat{k} \)
Next, find the magnitude of this cross product:
\( | \vec{a} \times \vec{b} | = \sqrt{6^2 + 5^2 + (-8)^2} \)
\( \implies | \vec{a} \times \vec{b} | = \sqrt{36 + 25 + 64} \)
\( \implies | \vec{a} \times \vec{b} | = \sqrt{125} \)
Now, calculate the area of the triangle:
\( \text{Area of triangle} = \frac{1}{2} | \vec{a} \times \vec{b} | \)
\( \implies \text{Area of triangle} = \frac{1}{2} \sqrt{125} \)
We can simplify \( \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \).
\( \implies \text{Area of triangle} = \frac{5\sqrt{5}}{2} \) square units.
In simple words: When you have two sides of a triangle given as vectors, you can find the area by first multiplying the vectors in a special way (cross product). Then, you find the length of this new vector and divide it by two. That final number is the area of the triangle.

🎯 Exam Tip: Remember that the area of a triangle formed by two vectors \( \vec{a} \) and \( \vec{b} \) is half the magnitude of their cross product, i.e., \( \frac{1}{2} |\vec{a} \times \vec{b}| \).