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

Let unit vector a have (a_1, a_2, a_3 ) components. a =a_1 i +a_2 j +a_3 k Since a is a unit vector, | a |=1 Also it is given that a makes angles 3 with i , 4 with j , and an acute angle with k . Then, we have: 3= a_1 | a | 12=a_1 [| a |=1] 4= a_2 | a | 1 2 =a_2 [| a |=1] Also, = a_3 | a | a_3= Now, |a|=1 a_1^2+a_2^2+a_3^2 =1 (12 )^2+ (1 2 )^2+ ^2 =1 14+12+ ^2 =1 34+ ^2 =1 ^2 =1-34=14 =12 = 3 a_3= 3=12 Hence, = 3 and the components of a are (12,1 2 ,1 2 ).

If a unit vector a makes an angle 3 with i , 4 with j and an acut…

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Question

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

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Answer

Let unit vector $\vec{\text{a}}$ have $\left(a_1, a_2, a_3\right)$ components.
$\Rightarrow\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
Since $\vec{\text{a}}$ is a unit vector, $|\vec{\text{a}}|=1$
Also it is given that $\vec{\text{a}}$ makes angles $\frac{\pi}3$ with $\hat{\text{i}}$, $\frac{\pi}4$ with $\hat{\text{j}}$, and an acute angle $\theta$ with $\hat{\text{k}}$.
Then, we have:
$\cos\frac{\pi}3=\frac{\text{a}_1}{|\vec{\text{a}}|}$
$\Rightarrow\ \frac{1}2=\text{a}_1$ $[|\vec{\text{a}}|=1]$
$\cos\frac{\pi}4=\frac{\text{a}_2}{|\vec{\text{a}}|}$
$\Rightarrow\ \frac{1}{\sqrt2}=\text{a}_2$ $[|\vec{\text{a}}|=1]$
Also, $\cos\theta=\frac{\text{a}_3}{|\vec{\text{a}}|}$
$\Rightarrow\ \text{a}_3=\cos\theta$
Now,
$|\text{a}|=1$
$\Rightarrow\ \sqrt{\text{a}_1^2+\text{a}_2^2+\text{a}_3^2}=1$
$\Rightarrow\Big(\frac{1}2\Big)^2+\Big(\frac{1}{\sqrt{2}}\Big)^2+\cos^2\theta=1$
$\Rightarrow\ \frac{1}4+\frac{1}2+\cos^2\theta=1$
$\Rightarrow\ \frac{3}4+\cos^2\theta=1$
$\Rightarrow\ \cos^2\theta=1-\frac{3}4=\frac{1}4$
$\Rightarrow\ \cos\theta=\frac{1}2$
$\Rightarrow\ \theta=\frac{\pi}3$
$\therefore\ \text{a}_3=\cos\frac{\pi}3=\frac{1}2$
Hence, $\theta=\frac{\pi}3$ and the components of $\vec{\text{a}}$ are $\Big(\frac{1}2,\frac{1}{\sqrt2},\frac{1}{\sqrt2}\Big)$.