If a unit vector a makes angles 3 with i, 4 with j and an acute angle with k, then find and hence, the components of a .

Let a =x i +y j +z k Given that it is a unit vector, So, | a |= x^ 2 +y^ 2 +z^ 2 =1 ......(i) Let , , be the angles a makes with i, , k respectively. Then, = a |a|| | =x 1 = 3 (Given) x= = 3 =1 2 = a j | a || j | = y 1 = 4 (Given) y= = 4 =1 2 = a k | a || k | = z 1 z= Putting value of x,y, z in equation (i) x^ 2 +y^ 2 +z^ 2 =1 x^ 2 +y^ 2 +z^ 2 =1 (Squaring both sides) (1 2 )^ 2 + (1 2 )^ 2 +( )^ 2 =1 1 4 +1 2 + ^ 2 =1 ^ 2 =1-3 4 =1 4 = 1 2 As should be acute, = ^ -1 1 2 = 3 z= = 3 =1 2 Components of a are the coefficients of , , k which are x=1 2 ; y=1 2 ; z=1 2 with = 3

If a unit vector a makes angles 3 with i, 4 with j and an acute a…

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Question

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 hence, the components of $\vec a $.

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Answer

Let $\vec{\mathrm{a}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$
Given that it is a unit vector,
So, $|\vec{\mathrm{a}}|=\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}}=1$ ......(i)
Let $\alpha, \beta, \theta$ be the angles $\vec a$ makes with $\hat{i}, \hat{\jmath}, \hat{\mathrm{k}}$ respectively.
Then,
$\cos \alpha=\frac{\vec{a} \cdot \hat{\imath}}{|\vec{a}||\hat{\imath}|}=\frac{x}{1}$
$\Rightarrow \alpha=\frac{\pi}{3}$ (Given)
$\Rightarrow x=\cos \alpha=\cos \frac{\pi}{3}=\frac{1}{2}$
$\cos \beta=\frac{\vec{\mathrm{a}} \cdot \hat{\mathrm{j}}}{|\vec{\mathrm{a}}||\hat{\mathrm{j}}|}=\frac{\mathrm{y}}{1}$
$\Rightarrow \beta=\frac{\pi}{4}$ (Given)
$\Rightarrow \mathrm{y}=\cos \beta=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
$\cos \theta=\frac{\vec{\mathrm{a}} \cdot \hat{\mathrm{k}}}{|\vec{\mathrm{a}}||\hat{\mathrm{k}}|}=\frac{\mathrm{z}}{1}$
$\Rightarrow {z}=\cos \theta$
Putting value of x,y, z in equation (i)
$\sqrt{x^{2}+y^{2}+z^{2}}=1$
$\Rightarrow x^{2}+y^{2}+z^{2}=1$ (Squaring both sides)
$\Rightarrow\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+(\cos \theta)^{2}=1$
$\Rightarrow \frac{1}{4}+\frac{1}{2}+\cos ^{2} \theta=1$
$\Rightarrow \cos ^{2} \theta=1-\frac{3}{4}=\frac{1}{4}$
$\Rightarrow \cos \theta=\pm \frac{1}{2}$
As $\theta$ should be acute, $\theta=\cos ^{-1} \frac{1}{2}=\frac{\pi}{3}$
$\therefore \mathrm{z}=\cos \theta=\cos \frac{\pi}{3}=\frac{1}{2}$
Components of $\vec{\mathrm{a}}$ are the coefficients of $\hat{\imath}, \hat{\jmath}, \hat{\mathrm{k}}$ which are
$x=\frac{1}{2} ; y=\frac{1}{\sqrt{2}} ; z=\frac{1}{2}$ with $\theta=\frac{\pi}{3}$

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