Vectors vec{a}, vec{b}, vec{c}, vec{d} are su | vedclass

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

Question

(A) Let $\vec{N_1}$ and $\vec{N_2}$ be the normal vectors to the planes $P_1$ and $P_2$ respectively. Since $P_1$ is determined by $\vec{a}$ and $\vec

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $\vec{N_1}$ and $\vec{N_2}$ be the normal vectors to the planes $P_1$ and $P_2$ respectively. Since $P_1$ is determined by $\vec{a}$ and $\vec{b}$,its normal is $\vec{N_1} = \vec{a} 	imes \vec{b}$. Since $P_2$ is determined by $\vec{c}$ and $\vec{d}$,its normal is $\vec{N_2} = \vec{c} 	imes \vec{d}$. Given that $(\vec{a} 	imes \vec{b}) 	imes (\vec{c} 	imes \vec{d}) = \vec{0}$,we have $\vec{N_1} 	imes \vec{N_2} = \vec{0}$. This implies that the normal vectors $\vec{N_1}$ and $\vec{N_2}$ are parallel to each other. Since the normal vectors are parallel,the planes $P_1$ and $P_2$ are parallel. Therefore,the angle between the planes $P_1$ and $P_2$ is $0$.

Vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are such that $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0}$. $P_1$ and $P_2$ are two planes determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}, \vec{d}$ respectively. Then the angle between the planes $P_1$ and $P_2$ is

Similar Questions

If the two diagonals of a parallelogram are $\bar{d_1} = \bar{i} + 2\bar{j} + 3\bar{k}$ and $\bar{d_2} = -2\bar{i} + \bar{j} - 2\bar{k}$,then the area of the parallelogram in square units is

The area of a parallelogram whose adjacent sides are given by the vectors $i + 2j + 3k$ and $-3i - 2j + k$ (in square units) is

$r \times a = b \times a;\,\,r \times b = a \times b;\,\,a \ne 0;\,\,b \ne 0;\,\,a \ne \lambda b;\,\,a$ is not perpendicular to $b,$ then $r = $

If $\overline{a}=\hat{i}+\hat{j}+\hat{k}$,$\overline{a} \cdot \overline{b}=1$ and $\overline{a} \times \overline{b}=\hat{j}-\hat{k}$,then $\overline{b}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module