A unit vector vec{a} makes an angle frac{pi}{ | vedclass

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(C) Let $\vec{a} = l\hat{i} + m\hat{j} + n\hat{k}$,where $l^2 + m^2 + n^2 = 1$.Since $\vec{a}$ makes an angle $\frac{\pi}{4}$ with the $z$-axis,$n = \

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$-\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$

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(C) Let $\vec{a} = l\hat{i} + m\hat{j} + n\hat{k}$,where $l^2 + m^2 + n^2 = 1$.Since $\vec{a}$ makes an angle $\frac{\pi}{4}$ with the $z$-axis,$n = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.Thus,$l^2 + m^2 + (\frac{1}{\sqrt{2}})^2 = 1 \implies l^2 + m^2 = \frac{1}{2} \dots (i)$.Given that $\vec{a} + \hat{i} + \hat{j}$ is a unit vector,we have $|(l+1)\hat{i} + (m+1)\hat{j} + \frac{1}{\sqrt{2}}\hat{k}| = 1$.Squaring both sides,$(l+1)^2 + (m+1)^2 + (\frac{1}{\sqrt{2}})^2 = 1^2$.$l^2 + 2l + 1 + m^2 + 2m + 1 + \frac{1}{2} = 1$.$(l^2 + m^2) + 2(l+m) + 2.5 = 1$.Substituting $(i)$,$\frac{1}{2} + 2(l+m) + 2.5 = 1 \implies 2(l+m) = -2 \implies l+m = -1$.Since $l^2 + m^2 = \frac{1}{2}$ and $l+m = -1$,we have $(l+m)^2 = l^2 + m^2 + 2lm = 1 \implies \frac{1}{2} + 2lm = 1 \implies 2lm = \frac{1}{2} \implies lm = \frac{1}{4}$.Solving $l+m = -1$ and $lm = \frac{1}{4}$ gives $l = m = -\frac{1}{2}$.Therefore,$\vec{a} = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$.

$A$ unit vector $\vec{a}$ makes an angle $\frac{\pi}{4}$ with the $z$-axis. If $\vec{a} + \hat{i} + \hat{j}$ is a unit vector,then $\vec{a}$ is equal to

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