Let a, b, c in mathbb{R} be such that a^{2} + | vedclass

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

Question

(D) Let $\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$.Given $a^{2} + b^{2} + c^{2} = 1$,we have $|\v

Vedclass · Accepted Answer

$\frac{2\pi}{3}$

Vedclass · Answer

(D) Let $\vec{p} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{q} = b \hat{i} + c \hat{j} + a \hat{k}$.Given $a^{2} + b^{2} + c^{2} = 1$,we have $|\vec{p}| = \sqrt{a^{2} + b^{2} + c^{2}} = 1$ and $|\vec{q}| = \sqrt{b^{2} + c^{2} + a^{2}} = 1$.The dot product is $\vec{p} \cdot \vec{q} = ab + bc + ca$.Let $a \cos 	heta = b \cos \left(	heta + \frac{2\pi}{3}ight) = c \cos \left(	heta + \frac{4\pi}{3}ight) = k$.Then $a = \frac{k}{\cos 	heta}$,$b = \frac{k}{\cos(	heta + 2\pi/3)}$,$c = \frac{k}{\cos(	heta + 4\pi/3)}$.Since $a+b+c = k(\sec 	heta + \sec(	heta + 2\pi/3) + \sec(	heta + 4\pi/3)) = 0$ (using the identity for sum of secants),we have $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) = 0$.Since $a^{2} + b^{2} + c^{2} = 1$,we get $1 + 2(ab + bc + ca) = 0$,so $ab + bc + ca = -1/2$.Thus,$\cos \phi = \frac{\vec{p} \cdot \vec{q}}{|\vec{p}||\vec{q}|} = \frac{-1/2}{1 	imes 1} = -1/2$.Therefore,$\phi = \frac{2\pi}{3}$.

Similar Questions

If the position vectors of $A$ and $B$ are $6i + j - 3k$ and $4i - 3j - 2k$ respectively,then the work done by the force $\vec{F} = i - 3j + 5k$ in displacing a particle from $A$ to $B$ is ............ $units$.

If $\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$,$|\vec{a}| = 1$,$|\vec{b}| = 2$,and $|\vec{c}| = 3$,then find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

What will be the length of the longer diagonal of the parallelogram constructed on vectors $5a + 2b$ and $a - 3b$,given that $|a| = 2\sqrt{2}$,$|b| = 3$,and the angle between $a$ and $b$ is $\frac{\pi}{4}$?

If $\overrightarrow{F_1} = i - j + k,$ $\overrightarrow{F_2} = -i + 2j - k,$ $\overrightarrow{F_3} = j - k,$ $\vec{A} = 4i - 3j - 2k$ and $\vec{B} = 6i + j - 3k,$ then the scalar product of $(\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3})$ and $\overrightarrow{AB}$ will be:

Let $a = 2i + j + k$ and $b = i + 2j - k$. If a unit vector $c$ is coplanar with $a$ and $b$,and $c$ is perpendicular to $a$,then $c$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module