Question

If a unit vector $\overrightarrow{u}=x\hat{i}+y\hat{j}+z\hat{k}$ which makes an angle of ${45}^{\circ }$ with the vector $2\hat{i}+2\hat{j}-\hat{k}$ and an angle of ${60}^{\circ }$ with the vector $\hat{j}-\hat{k}.$ Then point $P(x,y,z)$ is

• A. $(\frac{1}{3\sqrt{2}},\frac{4}{3\sqrt{2}},\frac{1}{3\sqrt{2}})$
• B. $(\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}})$
• C. $(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}})$
• D. $(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$

Answer 1

Answer: (A), (C)

$(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}})$

Let unit vector be $c=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ which makes an angle of ${45}^{\circ }$ with $2\hat{i}+2\hat{j}-\hat{k}$
$\therefore \frac{2c_1+2c_2-c_3}{3}=\cos {45}^{\circ }\Rightarrow 2c_1+2c_2-c_3=\frac{3}{\sqrt{2}}\cdots \left(i\right)$
Also it makes an angle ${60}^{\circ }$ with $\hat{j}-\hat{k}$
$\Rightarrow \frac{c_2-c_3}{\sqrt{2}}=\cos {60}^{\circ }\Rightarrow c_2-c_3=\frac{\sqrt{2}}{2}\cdots \left(ii\right)$
Also, $c_1^2+c_2^2+c_3^2=1\cdots \left(iii\right)$
Now, we have from equation $\left(ii\right),c_2=c_3+\frac{1}{\sqrt{2}}$ and from equation $\left(i\right),c_1=\frac{-c_3}{2}+\frac{1}{2\sqrt{2}}$
On substituting in equation $\left(iii\right)$, we get $\frac{c_3^2}{4}+\frac{1}{2}+\frac{2c_3}{\sqrt{2}}+\frac{1}{8}-\frac{c_3}{2\sqrt{2}}+c_3^2+c_3^2=1\Rightarrow \frac{9c_3^2}{4}+\frac{3c_3}{2\sqrt{2}}+\frac{5}{8}=1\Rightarrow c_3=-\frac{1}{\sqrt{2}},\frac{1}{3\sqrt{2}}$
Hence the required vectors are $(\frac{1}{3\sqrt{2}},\frac{4}{3\sqrt{2}},\frac{1}{3\sqrt{2}})$ and $(\frac{1}{\sqrt{2}},0,-\frac{1}{\sqrt{2}})$