A vector perpendicular to hat{i}+hat{j}+hat{k | vedclass

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(D) Two vectors $\vec{A}$ and $\vec{B}$ are perpendicular if their dot product is zero,i.e.,$\vec{A} \cdot \vec{B} = 0$.Let $\vec{A} = \hat{i} + \hat{

Vedclass · Accepted Answer

$3\hat{i}+2\hat{j}-5\hat{k}$

Vedclass · Answer

(D) Two vectors $\vec{A}$ and $\vec{B}$ are perpendicular if their dot product is zero,i.e.,$\vec{A} \cdot \vec{B} = 0$.Let $\vec{A} = \hat{i} + \hat{j} + \hat{k}$.For option $A$: $(\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1 
eq 0$.For option $B$: $(\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} - \hat{k}) = (1)(1) + (1)(-1) + (1)(-1) = 1 - 1 - 1 = -1 
eq 0$.For option $C$: $(\hat{i} + \hat{j} + \hat{k}) \cdot (-\hat{i} - \hat{j} - \hat{k}) = (1)(-1) + (1)(-1) + (1)(-1) = -1 - 1 - 1 = -3 
eq 0$.For option $D$: $(\hat{i} + \hat{j} + \hat{k}) \cdot (3\hat{i} + 2\hat{j} - 5\hat{k}) = (1)(3) + (1)(2) + (1)(-5) = 3 + 2 - 5 = 0$.Since the dot product is zero for option $D$,it is the correct answer.

$A$ vector perpendicular to $\hat{i}+\hat{j}+\hat{k}$ is:

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