If a non-zero vector vec{a} is parallel to th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The normal $\vec{n}_1$ to the plane $P_1$ determined by $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ is given by the cross product: $\vec{n}_1 = (\ha

Vedclass · Accepted Answer

$\cos^{-1}\left(\pm\frac{1}{\sqrt{3}}\right)$

Vedclass · Answer

(D) The normal $\vec{n}_1$ to the plane $P_1$ determined by $\hat{j}-\hat{k}$ and $3\hat{j}-2\hat{k}$ is given by the cross product: $\vec{n}_1 = (\hat{j}-\hat{k}) 	imes (3\hat{j}-2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -1 \ 0 & 3 & -2 \end{vmatrix} = \hat{i}(-2+3) = \hat{i}$. The normal $\vec{n}_2$ to the plane $P_2$ determined by $2\hat{i}+3\hat{j}$ and $\hat{i}-3\hat{j}$ is given by: $\vec{n}_2 = (2\hat{i}+3\hat{j}) 	imes (\hat{i}-3\hat{j}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 0 \ 1 & -3 & 0 \end{vmatrix} = \hat{k}(-6-3) = -9\hat{k}$. Since $\vec{a}$ is parallel to the line of intersection of planes $P_1$ and $P_2$,it is parallel to $\vec{n}_1 	imes \vec{n}_2$: $\vec{a} = k(\vec{n}_1 	imes \vec{n}_2) = k(\hat{i} 	imes -9\hat{k}) = k(9\hat{j}) = 9k\hat{j}$. We can take $\vec{a} = \pm\hat{j}$. The angle $	heta$ between $\vec{a} = \hat{j}$ and $\vec{b} = \hat{i}+\hat{j}+\hat{k}$ is given by: $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{1 \cdot \sqrt{1^2+1^2+1^2}} = \frac{1}{\sqrt{3}}$. Considering the direction $\vec{a} = -\hat{j}$,we get $\cos 	heta = -\frac{1}{\sqrt{3}}$. Thus,$	heta = \cos^{-1}\left(\pm\frac{1}{\sqrt{3}}ight)$.

Similar Questions

The area of a parallelogram whose two adjacent sides are represented by the vectors $\vec{a} = 3i - k$ and $\vec{b} = i + 2j$ is

If either $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0},$ then $\vec{a} \times \vec{b}=\overrightarrow{0} .$ Is the converse true? Justify your answer with an example.

Find the area of a parallelogram whose adjacent sides are given by the vectors $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

$A$ vector perpendicular to the plane containing the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ is

Let the vectors $a, b, c$ and $d$ be such that $(a \times b) \times (c \times d) = 0$. Let $P_1$ and $P_2$ be planes determined by the pairs of vectors $(a, b)$ and $(c, d)$ respectively. Then the angle between $P_1$ and $P_2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module