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GUJCET · Maths · Vector Algebra

A unit vector perpendicular to each of the vectors \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) is ___________ , where \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\)

  1. A \(-\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{j}-\frac{1}{\sqrt{6}} \hat{k}\)
  2. B \(-\frac{1}{\sqrt{12}} \hat{i}+\frac{2}{\sqrt{12}} \hat{j}-\frac{1}{\sqrt{12}} \hat{k}\)
  3. C \(\frac{1}{\sqrt{12}} \hat{i}+\frac{2}{\sqrt{12}} \hat{j}-\frac{1}{\sqrt{12}} \hat{k}\)
  4. D \(\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{j}-\frac{1}{\sqrt{6}} \hat{k}\)

Step-by-step Solution

Detailed explanation

\( (\vec{a}+\vec{b}) = (\hat{i}+\hat{j}+\hat{k}) + (\hat{i}+2 \hat{j}+3 \hat{k}) = 2\hat{i}+3\hat{j}+4\hat{k} \) \( (\vec{a}-\vec{b}) = (\hat{i}+\hat{j}+\hat{k}) - (\hat{i}+2 \hat{j}+3 \hat{k}) = -\hat{j}-2\hat{k} \)