A non-zero vector vec{a} is parallel to the l | vedclass

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

Question

(B) The equation of the plane containing vectors $\hat{i}$ and $\hat{i}+\hat{j}$ is given by the normal vector $\vec{n}_1 = \hat{i} 	imes (\hat{i}+\h

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) The equation of the plane containing vectors $\hat{i}$ and $\hat{i}+\hat{j}$ is given by the normal vector $\vec{n}_1 = \hat{i} 	imes (\hat{i}+\hat{j}) = \hat{k}$. Thus, the plane equation is $z = 0$.The equation of the plane containing vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$ is given by the normal vector $\vec{n}_2 = (\hat{i}-\hat{j}) 	imes (\hat{i}+\hat{k}) = \hat{i} 	imes \hat{k} - \hat{j} 	imes \hat{i} - \hat{j} 	imes \hat{k} = -\hat{j} + \hat{k} - \hat{i} = -\hat{i} - \hat{j} + \hat{k}$. Thus, the plane equation is $x + y - z = 0$.Since $\vec{a}$ is parallel to the line of intersection of these two planes, $\vec{a}$ must be parallel to the cross product of their normals: $\vec{v} = \vec{n}_1 	imes \vec{n}_2 = \hat{k} 	imes (-\hat{i} - \hat{j} + \hat{k}) = -(\hat{k} 	imes \hat{i}) - (\hat{k} 	imes \hat{j}) = -\hat{j} + \hat{i} = \hat{i} - \hat{j}$.Let $\vec{b} = \hat{i} - \hat{j}$. The angle $	heta$ between $\vec{b}$ and $\vec{c} = \hat{i} - 2\hat{j} + 2\hat{k}$ is given by $\cos 	heta = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{b}| |\vec{c}|}$.$\vec{b} \cdot \vec{c} = (1)(1) + (-1)(-2) + (0)(2) = 1 + 2 = 3$.$|\vec{b}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.$|\vec{c}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1+4+4} = 3$.$\cos 	heta = \frac{3}{\sqrt{2} 	imes 3} = \frac{1}{\sqrt{2}}$.Therefore, $	heta = \frac{\pi}{4}$.

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}$ and $\hat{i}+\hat{j}$ and the plane determined by vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{k}$. The angle between $\vec{a}$ and $(\hat{i}-2\hat{j}+2\hat{k})$ is

Similar Questions

One side and one diagonal of a parallelogram are represented by $3 \hat{i}+\hat{j}-\hat{k}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ respectively. Then,the area of the parallelogram in square units is:

For vectors $\vec{a}$ and $\vec{b}$,if $|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector,then the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k$,then find the unit vector perpendicular to both $u$ and $v$.

If $u = 2i + 2j - k$ and $v = 6i - 3j + 2k,$ then a unit vector perpendicular to both $u$ and $v$ is

If the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ are coplanar,find the unit vector perpendicular to the plane containing these points.

Vedclass Test Series

Exam Paper Generator

Online Exam Module