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  • The number of vectors of unit length perpendicular to the vectors a =2i +j +2k and b =j +k is A. one

The number of vectors of unit length perpendicular to the vectors \vec{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\text { and } \vec{\mathrm{b}}=\hat{\mathrm{j}}+\hat{\mathrm{k}}  is
A. one
B. two
C. three
D. infinite

Answers (1)

Answer :(B)

Given that ,   \vec{\mathrm{a}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\text { and } \vec{\mathrm{b}}=\hat{\mathrm{j}}+\hat{\mathrm{k}} 

Now, a vector which is perpendicular to both \vec{a} \text{ and } \vec{b} is given by

\\ \vec{a} \times \vec{b}=\left|\begin{array}{lll}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1\end{array}\right|=\hat{\imath}(1-2)-\hat{\jmath}(2-0)+\hat{\mathrm{k}}(2-0)=-\hat{\imath}-2 \hat{\jmath}+2 \hat{\mathrm{k}}$\\ Now, $|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-2)^{2}+(2)^{2}}=\sqrt{1+4+4}=\sqrt{9}=3$ \\$\therefore$ the required unit vector

\\ =\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{-\hat{i}-2 \hat{\jmath}+2 \hat{k}}{3}=\frac{-1}{3} \hat{\imath}-\frac{2}{3} \hat{\jmath}+\frac{2}{3} k$\\ There are two perpendicular directions to any plane. Thus, another unit vector perpendicular to $\vec{a}$ and $\vec{b}$ is given $b y-\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}$\\ $\Rightarrow \frac{\vec{b} \times \vec{a}}{|\vec{b} \times \vec{a}|}=\frac{1}{3} \hat{\imath}+\frac{2}{3} \hat{\jmath}-\frac{2}{3} k$

Hence, there are two unit length perpendicular to the \vec{a} \text{ and } \vec{b} .

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