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A unit vector a makes an angle $\frac{\pi}{4}$ with z-axis. If $\mathbf{a}+\mathbf{i}+\mathbf{j}$ is a unit vector, then $\mathbf{a}$ is equal to
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The correct answer is:
$-\frac{\mathbf{i}}{2}-\frac{\mathbf{j}}{2}+\frac{\mathbf{k}}{\sqrt{2}}$
Let $\mathrm{a}=l \hat{i}+m \hat{j}+n \hat{k}$, where $l^2+m^2+n^2=1$.
a makes an angle $\frac{\pi}{4}$ with $z-$ axis.
$\begin{aligned}
& \therefore n=\frac{1}{\sqrt{2}}, \quad I^2+m^2=\frac{1}{2} \\
& \therefore \mathrm{a}=l \mathrm{i}+m \mathrm{j}+\frac{\mathrm{k}}{\sqrt{2}} \\
& \mathrm{a}+\mathrm{i}+\mathrm{j}=(l+1) \mathrm{i}+(m+1) \mathrm{j}+\frac{\mathrm{k}}{\sqrt{2}} \\
& \text {Its magnitude is } 1 \text {, hence }(l+1)^2+(m+1)^2=\frac{1}{2}
\end{aligned}$
Its magnitude is 1, hence
$(l+1)^2+(m+1)^2=\frac{1}{2}$
From (i) and (ii),
$2 I m=\frac{1}{2} \Rightarrow I=m=-\frac{1}{2}$
Hence
$a=-\frac{\mathrm{i}}{2}-\frac{\mathrm{j}}{2}+\frac{\mathrm{k}}{\sqrt{2}}$
a makes an angle $\frac{\pi}{4}$ with $z-$ axis.
$\begin{aligned}
& \therefore n=\frac{1}{\sqrt{2}}, \quad I^2+m^2=\frac{1}{2} \\
& \therefore \mathrm{a}=l \mathrm{i}+m \mathrm{j}+\frac{\mathrm{k}}{\sqrt{2}} \\
& \mathrm{a}+\mathrm{i}+\mathrm{j}=(l+1) \mathrm{i}+(m+1) \mathrm{j}+\frac{\mathrm{k}}{\sqrt{2}} \\
& \text {Its magnitude is } 1 \text {, hence }(l+1)^2+(m+1)^2=\frac{1}{2}
\end{aligned}$
Its magnitude is 1, hence
$(l+1)^2+(m+1)^2=\frac{1}{2}$
From (i) and (ii),
$2 I m=\frac{1}{2} \Rightarrow I=m=-\frac{1}{2}$
Hence
$a=-\frac{\mathrm{i}}{2}-\frac{\mathrm{j}}{2}+\frac{\mathrm{k}}{\sqrt{2}}$
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