MHT CET · Maths · Vector Algebra
The number of unit vectors perpendicular to \(\overline{\mathrm{a}}=(1,1,0)\) and \(\overline{\mathrm{b}}=(0,1,1)\) is
- A one.
- B two.
- C three.
- D infinite.
Answer & Solution
Correct Answer
(B) two.
Step-by-step Solution
Detailed explanation
The vector perpendicular to \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) is
\(\bar{a} \times \bar{b}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 0 \\
0 & 1 & 0
\end{array}\right|=\hat{i}-\hat{j}+\hat{k}\)
Since the length of this vector is \(\sqrt{3}\), the unit vector perpendicular to \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) is \(\pm \frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}= \pm \frac{1}{\sqrt{3}}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
Hence, there are two such vectors.
\(\bar{a} \times \bar{b}=\left|\begin{array}{lll}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 0 \\
0 & 1 & 0
\end{array}\right|=\hat{i}-\hat{j}+\hat{k}\)
Since the length of this vector is \(\sqrt{3}\), the unit vector perpendicular to \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{b}}\) is \(\pm \frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}= \pm \frac{1}{\sqrt{3}}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
Hence, there are two such vectors.
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