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

Solution

✅ Correct Option: 1

A unit vector perpendicular to both i^−j^\hat{i} - \hat{j} and i^+j^\hat{i} + \hat{j} can be found using the cross product, which gives a vector perpendicular to both input vectors.


The vectors in 3D form:

a⃗=i^−j^+0k^\vec{a} = \hat{i} - \hat{j} + 0\hat{k}

b⃗=i^+j^+0k^\vec{b} = \hat{i} + \hat{j} + 0\hat{k}


The cross product:

a⃗×b⃗=∣i^j^k^1−10110∣\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 0 \end{vmatrix}

=i^∣−1010∣−j^∣1010∣+k^∣1−111∣= \hat{i}\begin{vmatrix} -1 & 0 \\ 1 & 0 \end{vmatrix} - \hat{j}\begin{vmatrix} 1 & 0 \\ 1 & 0 \end{vmatrix} + \hat{k}\begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}

=i^[(−1)(0)−(0)(1)]−j^[(1)(0)−(0)(1)]+k^[(1)(1)−(−1)(1)]= \hat{i}[(-1)(0) - (0)(1)] - \hat{j}[(1)(0) - (0)(1)] + \hat{k}[(1)(1) - (-1)(1)]

=i^(0)−j^(0)+k^(2)= \hat{i}(0) - \hat{j}(0) + \hat{k}(2)

=2k^= 2\hat{k}


The magnitude of 2k^2\hat{k}:

∣a⃗×b⃗∣=02+02+22|\vec{a} \times \vec{b}| = \sqrt{0^2 + 0^2 + 2^2}

=2= 2


The unit vector:

n^=2k^2\hat{n} = \frac{2\hat{k}}{2}

=k^= \hat{k}

Both given vectors lie in the xy-plane. Any vector perpendicular to this plane points in the z-direction.

Therefore, the unit vector perpendicular to both given vectors is k^\hat{k}.

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