Question

A unit vector $\vec{a}$ in the plane of $\vec{b} = 2\hat{i} + \hat{j}$ & $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ is such that $\vec{a} \land \vec{b} = \vec{a} \land \vec{d}$ where $\vec{d} = \hat{j} + 2\hat{k}$ is

• A. $\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$
• B. $\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$
• C. $\frac{2\hat{i}+\hat{j}}{\sqrt{5}}$
• D. $\frac{2\hat{i}+\hat{j}}{\sqrt{5}}$

Answer 1

Answer: (B)

$\displaystyle \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$

Answer 2

We shall show that if we interpret “∧” as the dot product (so that the condition is

$$\vec a\cdot\vec b=\vec a\cdot\vec d,$$

) then the answer turns out to be option (2).

Let

$$\vec b=2\hat{i}+\hat{j},\quad \vec d=\hat{j}+2\hat{k},$$

and suppose that the required unit vector $\vec a=(x,y,z)$ lies in the plane of $\vec b$ and $\vec c=\hat{i}-\hat{j}+\hat{k}$. (In other words, $\vec a$ can be expressed as a linear combination of $\vec b$ and $\vec c$.)

The condition

$$\vec a\cdot\vec b=\vec a\cdot\vec d$$

implies

$$2x+y = y+2z\quad\Longrightarrow\quad 2x=2z,\quad\text{or}\quad x=z.$$

Also, since $\vec a$ lies in the plane of $\vec b$ and $\vec c$, it must be perpendicular to the normal to that plane. One may quickly show that

$$\vec n=\vec b\times\vec c=(1,-2,-3).$$

Thus

$$\vec a\cdot\vec n=x-2y-3z=0.$$

But since $x=z$ this becomes

$$x-2y-3x=-2y-2x=0\quad\Longrightarrow\quad y=-x.$$

Thus we have

$$x=z,\quad y=-x.$$

A unit vector (with $x\ne 0$) is obtained by choosing

$$x=\frac{1}{\sqrt{3}},$$

so that

$$\vec a=\frac{1}{\sqrt{3}}(1,-1,1).$$

Checking the dot‐products,

$$\vec a\cdot\vec b=\frac{1}{\sqrt{3}}(1\cdot2+(-1)\cdot1+1\cdot0)=\frac{2-1}{\sqrt{3}}=\frac{1}{\sqrt{3}},$$

and

$$\vec a\cdot\vec d=\frac{1}{\sqrt{3}}(1\cdot0+(-1)\cdot1+1\cdot2)=\frac{0-1+2}{\sqrt{3}}=\frac{1}{\sqrt{3}},$$

so the condition is satisfied.

Thus the answer is option (2).