The unit vector which is orthogonal to the ve | vedclass

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(C) Let the vector be $\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$.Since $\vec{v}$ is coplanar with $\vec{a} = 2 \hat{i} + \hat{j} + \hat{k}$ and $\v

Vedclass · Accepted Answer

$\frac{-8 \hat{i}+3 \hat{j}+3 \hat{k}}{\sqrt{82}}$

Vedclass · Answer

(C) Let the vector be $\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$.Since $\vec{v}$ is coplanar with $\vec{a} = 2 \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$,it must be of the form $\vec{v} = \lambda \vec{a} + \mu \vec{b} = (2\lambda + \mu) \hat{i} + (\lambda + \mu) \hat{j} + (\lambda + \mu) \hat{k}$.Since $\vec{v}$ is orthogonal to $\vec{c} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$,we have $\vec{v} \cdot \vec{c} = 0$.$3(2\lambda + \mu) + 2(\lambda + \mu) + 6(\lambda + \mu) = 0$.$6\lambda + 3\mu + 2\lambda + 2\mu + 6\lambda + 6\mu = 14\lambda + 11\mu = 0$.Let $\lambda = 11$,then $\mu = -14$.$\vec{v} = (2(11) - 14) \hat{i} + (11 - 14) \hat{j} + (11 - 14) \hat{k} = 8 \hat{i} - 3 \hat{j} - 3 \hat{k}$.The unit vector is $\pm \frac{8 \hat{i} - 3 \hat{j} - 3 \hat{k}}{\sqrt{64 + 9 + 9}} = \pm \frac{8 \hat{i} - 3 \hat{j} - 3 \hat{k}}{\sqrt{82}}$.Comparing with the options,the correct choice is $\frac{-8 \hat{i} + 3 \hat{j} + 3 \hat{k}}{\sqrt{82}}$ which is option $C$.

The unit vector which is orthogonal to the vector $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and coplanar with the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+\hat{k}$ is

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