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Mathematics Question on Vector Algebra

The unit vector perpendicular to each of the vectors a⃗+b⃗ and a⃗−b⃗, where a⃗=i^+j^+k^ and b⃗=i^+2j^+3k^, is:\text{The unit vector perpendicular to each of the vectors } \vec{a} + \vec{b} \text{ and } \vec{a} - \vec{b}, \text{ where } \vec{a} = \hat{i} + \hat{j} + \hat{k} \text{ and } \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}, \text{ is:}

A

16i^+26j^+16k^\frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}

B

−16i^+16j^−16k^-\frac{1}{\sqrt{6}} \hat{i} + \frac{1}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k}

C

−16i^+26j^+26k^-\frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} + \frac{2}{\sqrt{6}} \hat{k}

D

−16i^+26j^−16k^-\frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k}

Answer

−16i^+26j^−16k^-\frac{1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k}

Explanation

Solution

Let the given vectors be:
a⃗=i^+j^+k^,b⃗=i^+2j^+3k^.\vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}. We need to find the unit vector perpendicular to both a⃗+b⃗\vec{a} + \vec{b} and a⃗−b⃗.\vec{a} - \vec{b}. The cross product of these two vectors will give a vector perpendicular to both. Let’s first compute the cross product of a⃗+b⃗\vec{a} + \vec{b} and a⃗−b⃗.\vec{a} - \vec{b}.

First, compute the two vectors:
a⃗+b⃗=(i^+j^+k^)+(i^+2j^+3k^)=2i^+3j^+4k^,\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},
a⃗−b⃗=(i^+j^+k^)−(i^+2j^+3k^)=0i^−j^−2k^.\vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = 0\hat{i} - \hat{j} - 2\hat{k}.

Now compute the cross product:
(a⃗+b⃗)×(a⃗−b⃗)=∣i^j^k^ 234 0−1−2∣.(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 2 & 3 & 4 \\\ 0 & -1 & -2 \end{vmatrix}.

Expand the determinant:
(a⃗+b⃗)×(a⃗−b⃗)=i^∣34 −1−2∣−j^∣24 0−2∣+k^∣23 0−1∣.(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \hat{i} \begin{vmatrix} 3 & 4 \\\ -1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\\ 0 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\\ 0 & -1 \end{vmatrix}.

Calculate the 2x2 determinants:
∣34 −1−2∣=(3)(−2)−(4)(−1)=−6+4=−2,\begin{vmatrix} 3 & 4 \\\ -1 & -2 \end{vmatrix} = (3)(-2) - (4)(-1) = -6 + 4 = -2,
∣24 0−2∣=(2)(−2)−(4)(0)=−4,\begin{vmatrix} 2 & 4 \\\ 0 & -2 \end{vmatrix} = (2)(-2) - (4)(0) = -4,
∣23 0−1∣=(2)(−1)−(3)(0)=−2.\begin{vmatrix} 2 & 3 \\\ 0 & -1 \end{vmatrix} = (2)(-1) - (3)(0) = -2.

So the cross product is:
(a⃗+b⃗)×(a⃗−b⃗)=−2i^+4j^−2k^.(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = -2\hat{i} + 4\hat{j} - 2\hat{k}.

Now, to find the unit vector, we need to divide this vector by its magnitude.

The magnitude of the vector is:
∣∣v⃗∣∣=(−2)2+42+(−2)2=4+16+4=24=26.||\vec{v}|| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}.

Thus, the unit vector is:
v^=−2i^+4j^−2k^26=−16i^+26j^−16k^.\hat{v} = \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \frac{-1}{\sqrt{6}} \hat{i} + \frac{2}{\sqrt{6}} \hat{j} - \frac{1}{\sqrt{6}} \hat{k}.

This matches Option (4), so the correct answer is:
(4).