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Q1:
Find $|\vec{a} \times \vec{b}|$, if $\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$.
Solution :
Step 1: Setting up the Cross Product Matrix
To find the cross product $\vec{a} \times \vec{b}$, we utilize the determinant form of the cross product for vectors in three-dimensional Cartesian space [Per the definition of the vector cross product]:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Given $\vec{a} = 1\hat{i} - 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}$, we substitute the components:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}$
Step 2: Expanding the Determinant
Expanding along the first row [By Laplace expansion]:
$\vec{a} \times \vec{b} = \hat{i}[(-7)(2) - (7)(-2)] - \hat{j}[(1)(2) - (7)(3)] + \hat{k}[(1)(-2) - (-7)(3)]$
$\vec{a} \times \vec{b} = \hat{i}[-14 + 14] - \hat{j}[2 - 21] + \hat{k}[-2 + 21]$
$\vec{a} \times \vec{b} = 0\hat{i} - (-19)\hat{j} + 19\hat{k}$
$\vec{a} \times \vec{b} = 19\hat{j} + 19\hat{k}$
Step 3: Calculating the Magnitude
The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is defined as $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$. Applying this to our result:
$|\vec{a} \times \vec{b}| = \sqrt{(0)^2 + (19)^2 + (19)^2}$
$|\vec{a} \times \vec{b}| = \sqrt{0 + 361 + 361}$
$|\vec{a} \times \vec{b}| = \sqrt{722}$
Step 4: Simplifying the Radical
Factoring $722$:
$722 = 2 \times 361 = 2 \times 19^2$
$|\vec{a} \times \vec{b}| = \sqrt{19^2 \times 2} = 19\sqrt{2}$
Final Solution: $|\vec{a} \times \vec{b}| = 19\sqrt{2}$
More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.4
- 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 $\vec{b}$?
- 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 $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.
- 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
Chapters in CBSE - Class 12 Mathematics
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