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Mathematics Question on Vector Algebra

A unit vector perpendicular to both the vectors i^+j^\hat {i} +\hat { j} and j^+k^ \hat {j} +\hat {k} is

A

−i^−j^+k^3\frac {-\hat {i}-\hat{j}+\hat {k}} {\sqrt {3}}

B

i^+j^−k^3\frac {\hat {i}+\hat{j}-\hat {k}} { {3}}

C

i^+j^+k^3\frac {\hat {i}+\hat{j}+\hat {k}} {\sqrt {3}}

D

i^−j^+k^3\frac {\hat {i}-\hat{j}+\hat {k}} {\sqrt {3}}

Answer

i^−j^+k^3\frac {\hat {i}-\hat{j}+\hat {k}} {\sqrt {3}}

Explanation

Solution

Let a⃗=i^+j^\vec{a} = \hat{i} + \hat{j} and b⃗=j^+k^\vec{b} = \hat{j} +\hat{k}
Now, a⃗×b⃗=∣i^j^k^ 110 011∣\vec{a} \times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 1&1&0\\\ 0&1&1\end{vmatrix}
=i^(1−0)−j^(1−0)+k^(1−0)=\hat{i} \left(1-0\right) -\hat{j} \left(1-0\right) + \hat{k} \left(1-0\right)
=i^−j^+k^=\hat{i} -\hat{j} +\hat{k}
and ∣a⃗×b⃗∣=12+(−1)2+12=3\left|\vec{a} \times\vec{b}\right| = \sqrt{1^{2} +\left(-1\right)^{2} +1^{2}} = \sqrt{3}
∴\therefore Required unit vector =a⃗×b⃗∣a⃗×b⃗∣= \frac{\vec{a} \times\vec{b}}{\left|\vec{a} \times\vec{b}\right|}
=i^−j^+k^3= \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}