Unit vectors a, b and c are coplanar. A unit | vedclass

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

Question

(D) Since $a, b, c$ are coplanar,their scalar triple product $[a, b, c] = 0$.Given the vector triple product identity $(a 	imes b) 	imes (c 	imes d

Vedclass · Accepted Answer

$A$ and $C$

Vedclass · Answer

(D) Since $a, b, c$ are coplanar,their scalar triple product $[a, b, c] = 0$.Given the vector triple product identity $(a 	imes b) 	imes (c 	imes d) = [(a 	imes b) \cdot d]c - [(a 	imes b) \cdot c]d$.Since $c$ is coplanar with $a$ and $b$,$[a, b, c] = (a 	imes b) \cdot c = 0$,so the second term vanishes.Thus,$[(a 	imes b) \cdot d]c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$.Let $n$ be the unit vector perpendicular to $a$ and $b$. Then $a 	imes b = |a||b| \sin(30^\circ) n = (1)(1)(\frac{1}{2})n = \frac{1}{2}n$.Substituting this,we get $\frac{1}{2}(n \cdot d)c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$.Since $n$ and $d$ are both unit vectors perpendicular to the plane of $a, b, c$,$n \cdot d = \pm 1$.If $n \cdot d = 1$,then $\frac{1}{2}c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k \Rightarrow c = \frac{1}{3}i - \frac{2}{3}j + \frac{2}{3}k = \frac{i - 2j + 2k}{3}$.If $n \cdot d = -1$,then $-\frac{1}{2}c = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k \Rightarrow c = -\frac{1}{3}i + \frac{2}{3}j - \frac{2}{3}k = \frac{-i + 2j - 2k}{3}$.Both options $A$ and $C$ are valid,so the correct option is $D$.

Unit vectors $a, b$ and $c$ are coplanar. $A$ unit vector $d$ is perpendicular to them. If $(a \times b) \times (c \times d) = \frac{1}{6}i - \frac{1}{3}j + \frac{1}{3}k$ and the angle between $a$ and $b$ is $30^\circ$,then $c$ is:

Similar Questions

The sum of the distinct real values of $\mu$,for which the vectors $\mu \hat{i} + \hat{j} + \hat{k}$,$\hat{i} + \mu \hat{j} + \hat{k}$,and $\hat{i} + \hat{j} + \mu \hat{k}$ are coplanar,is

If $(1,5,35), (7,5,5), (1, \lambda, 7)$ and $(2 \lambda, 1, 2)$ are coplanar,then the sum of all possible values of $\lambda$ is

The volume of the tetrahedron having the edges $\hat{i}+2\hat{j}-\hat{k}$, $\hat{i}+\hat{j}+\hat{k}$, and $\hat{i}-\hat{j}+\lambda\hat{k}$ as coterminous edges is $\frac{2}{3}$ cubic units. Then $\lambda$ equals:

Let $\overrightarrow{A} = \hat{i} + \hat{j} + \hat{k}$,$\overrightarrow{B} = \hat{i}$,and $\overrightarrow{C} = C_1\hat{i} + C_2\hat{j} + C_3\hat{k}$. If $C_2 = -1$ and $C_3 = 1$,then to make the three vectors coplanar:

If the volume of a tetrahedron having $\bar{i}+2 \bar{j}-3 \bar{k}$, $2 \bar{i}+\bar{j}-3 \bar{k}$, and $3 \bar{i}-\bar{j}+p \bar{k}$ as its coterminous edges is $2$, then the values of $p$ are the roots of the equation

Vedclass Test Series

Exam Paper Generator

Online Exam Module