A unit vector coplanar with i+j+3k and i+3j+k | vedclass

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(C) Let the unit vector be $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.Let $\vec{a} = \hat{i} + \hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} + \h

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$\frac{1}{\sqrt{2}}(j-k)$

Vedclass · Answer

(C) Let the unit vector be $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.Let $\vec{a} = \hat{i} + \hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$, and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$.Since $\vec{r}$ is coplanar with $\vec{a}$ and $\vec{b}$, the scalar triple product $[\vec{r}, \vec{a}, \vec{b}] = 0$.$\left|\begin{matrix} x & y & z \ 1 & 1 & 3 \ 1 & 3 & 1 \end{matrix}ight| = 0 \Rightarrow x(1-9) - y(1-3) + z(3-1) = 0 \Rightarrow -8x + 2y + 2z = 0 \Rightarrow -4x + y + z = 0$ $(i)$.Since $\vec{r}$ is perpendicular to $\vec{c}$, $\vec{r} \cdot \vec{c} = 0$.$x + y + z = 0$ (ii).Subtracting $(i)$ from (ii): $(x + y + z) - (-4x + y + z) = 0 \Rightarrow 5x = 0 \Rightarrow x = 0$.Substituting $x=0$ into (ii), we get $y + z = 0 \Rightarrow y = -z$.Since $\vec{r}$ is a unit vector, $x^2 + y^2 + z^2 = 1$.$0^2 + y^2 + (-y)^2 = 1 \Rightarrow 2y^2 = 1 \Rightarrow y = \pm \frac{1}{\sqrt{2}}$.If $y = \frac{1}{\sqrt{2}}$, then $z = -\frac{1}{\sqrt{2}}$. If $y = -\frac{1}{\sqrt{2}}$, then $z = \frac{1}{\sqrt{2}}$.Thus, $\vec{r} = \pm \frac{1}{\sqrt{2}}(\hat{j} - \hat{k})$. Option $(C)$ is $\frac{1}{\sqrt{2}}(\hat{j} - \hat{k})$.

$A$ unit vector coplanar with $i+j+3k$ and $i+3j+k$ and perpendicular to $i+j+k$ is

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