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Q7:
Let the vectors $\vec{a}, \vec{b}, \vec{c}$ be given as $a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, $c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Then show that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.
Solution :
Step 1: Establishing Vector Components
Given the vectors in their component form:
- $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$
- $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$
- $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$
Step 2: Evaluating the Left-Hand Side (LHS)
First, calculate the sum of vectors $\vec{b}$ and $\vec{c}$ [Per the definition of Vector Addition]:
$\vec{b} + \vec{c} = (b_1 + c_1)\hat{i} + (b_2 + c_2)\hat{j} + (b_3 + c_3)\hat{k}$
Now, compute the cross product $\vec{a} \times (\vec{b} + \vec{c})$ using the determinant expansion [Per the definition of the Cross Product in Cartesian components]:
$\vec{a} \times (\vec{b} + \vec{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ (b_1+c_1) & (b_2+c_2) & (b_3+c_3) \end{vmatrix}$
Expanding along the first row:
$= \hat{i}[a_2(b_3+c_3) - a_3(b_2+c_2)] - \hat{j}[a_1(b_3+c_3) - a_3(b_1+c_1)] + \hat{k}[a_1(b_2+c_2) - a_2(b_1+c_1)]$
$= \hat{i}[a_2b_3 + a_2c_3 - a_3b_2 - a_3c_2] - \hat{j}[a_1b_3 + a_1c_3 - a_3b_1 - a_3c_1] + \hat{k}[a_1b_2 + a_1c_2 - a_2b_1 - a_2c_1]$
Step 3: Evaluating the Right-Hand Side (RHS)
Calculate $\vec{a} \times \vec{b}$ and $\vec{a} \times \vec{c}$ individually:
$\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\hat{i} + (a_3b_1 - a_1b_3)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$
$\vec{a} \times \vec{c} = (a_2c_3 - a_3c_2)\hat{i} + (a_3c_1 - a_1c_3)\hat{j} + (a_1c_2 - a_2c_1)\hat{k}$
Adding the two resultant vectors:
$\vec{a} \times \vec{b} + \vec{a} \times \vec{c} = [(a_2b_3 - a_3b_2) + (a_2c_3 - a_3c_2)]\hat{i} + [(a_3b_1 - a_1b_3) + (a_3c_1 - a_1c_3)]\hat{j} + [(a_1b_2 - a_2b_1) + (a_1c_2 - a_2c_1)]\hat{k}$
Step 4: Comparative Analysis
By regrouping the terms in the RHS expression, we obtain:
$= [(a_2b_3 + a_2c_3) - (a_3b_2 + a_3c_2)]\hat{i} + [(a_3b_1 + a_3c_1) - (a_1b_3 + a_1c_3)]\hat{j} + [(a_1b_2 + a_1c_2) - (a_2b_1 + a_2c_1)]\hat{k}$
Comparing this term-for-term with the expanded LHS in Step 2, we observe that the coefficients of $\hat{i}, \hat{j},$ and $\hat{k}$ are identical.
Final Solution: Since the components of $\vec{a} \times (\vec{b} + \vec{c})$ and $\vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ are identical, it is proven that $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.
More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.4
- Q1: Find $|\vec{a} \times \vec{b}|$, if $\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$.
- Q2: Find a unit vector perpendicular to each of the vector $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$.
- Q3: If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$, $\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, then find $\theta$ and hence, the components of $\vec{a}$.
- Q4: Show that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$
- Q5: Find $\lambda$ and $\mu$ if $(2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}$.
- Q6: Given that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?
- Q8: If either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \times \vec{b} = \vec{0}$. Is the converse true? Justify your answer with an example.
- Q9: Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
- Q10: Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} - 7\hat{j} + \hat{k}$.
- Q11: Let the vectors $\vec{a}$ and $\vec{b}$ be such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$, then $\vec{a} \times \vec{b}$ is a unit vector, if the angle between $\vec{a}$ and $\vec{b}$ is
- Q12: Area of a rectangle having vertices A, B, C and D with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$, $\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$, respectively is
CBSE Solutions for Class 12 Mathematics Vector Algebra
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