If a unit vector vec{a} makes angles frac{pi} | vedclass

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

Question

(A) Let the unit vector $\vec{a}$ be $\vec{a} = a_{1}\hat{i} + a_{2}\hat{j} + a_{3}\hat{k}$.Since $\vec{a}$ is a unit vector,$|\vec{a}| = 1$.The direc

Vedclass · Accepted Answer

$	heta = \frac{\pi}{3}, \vec{a} = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}ight)$

Vedclass · Answer

(A) Let the unit vector $\vec{a}$ be $\vec{a} = a_{1}\hat{i} + a_{2}\hat{j} + a_{3}\hat{k}$.Since $\vec{a}$ is a unit vector,$|\vec{a}| = 1$.The direction cosines of $\vec{a}$ are $\cos \alpha, \cos \beta, \cos \gamma$,where $\alpha = \frac{\pi}{3}, \beta = \frac{\pi}{4}$,and $\gamma = 	heta$.Thus,$a_{1} = \cos \frac{\pi}{3} = \frac{1}{2}$,$a_{2} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,and $a_{3} = \cos 	heta$.We know that $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$.Substituting the values: $\left(\frac{1}{2}ight)^{2} + \left(\frac{1}{\sqrt{2}}ight)^{2} + \cos^{2} 	heta = 1$.$\frac{1}{4} + \frac{1}{2} + \cos^{2} 	heta = 1$.$\frac{3}{4} + \cos^{2} 	heta = 1 \Rightarrow \cos^{2} 	heta = \frac{1}{4}$.Since $	heta$ is acute,$\cos 	heta = \frac{1}{2}$,which implies $	heta = \frac{\pi}{3}$.Therefore,the components of $\vec{a}$ are $\left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}ight)$.

If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$,$\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$,then find $\theta$ and the components of $\vec{a}$.

Similar Questions

If $A(a)$, $B(b)$, and $C(c)$ are vertices of $\Delta ABC$. Point $D$ divides segment $BC$ internally in the ratio $2 : 1$. Point $E$ divides segment $AD$ internally in the ratio $1 : 2$, then the position vector of $E$ is

The number of unit vectors of the form $a \hat{i} + b \hat{j} + c \hat{k}$,where $a, b, c \in W$ is

If three points $A$,$B$,and $C$ have position vectors $(1, x, 3)$,$(3, 4, 7)$,and $(y, -2, -5)$ respectively and if they are collinear,then $(x, y)$ is

If $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k},$ find the unit vector in the direction of
$(i)$ $6 \vec{b}$
(ii) $2 \vec{a}-\vec{b}$

In a regular hexagon $ABCDEF$,$AD + EB + FC = (3\lambda - 8) AB$. Then $\lambda =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module