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Question: The unit vector which is orthogonal to the vector $5\hat{i} + 2\hat{j} + 6\hat{k}$ and is coplanar w...

The unit vector which is orthogonal to the vector 5i^+2j^+6k^5\hat{i} + 2\hat{j} + 6\hat{k} and is coplanar with the vectors 2i^+j^+k^2\hat{i} + \hat{j} + \hat{k} and i^−j^+k^\hat{i} - \hat{j} + \hat{k} is

A

2i^−6j^+k^41\frac{2\hat{i} - 6\hat{j} + \hat{k}}{\sqrt{41}}

B

2i^−5j^29\frac{2\hat{i} - 5\hat{j}}{\sqrt{29}}

C

−3j^+k^10\frac{-3\hat{j} + \hat{k}}{\sqrt{10}}

D

2i^−8j^+k^69\frac{2\hat{i} - 8\hat{j} + \hat{k}}{69}

Answer

−3j^+k^10\frac{-3\hat{j} + \hat{k}}{\sqrt{10}}

Explanation

Solution

Solution:

Let the required unit vector be

u⃗=α(2,1,1)+β(1,−1,1)=(2α+β,  α−β,  α+β).\vec{u} = \alpha(2,1,1) + \beta(1,-1,1) = (2\alpha+\beta,\;\alpha-\beta,\;\alpha+\beta).

The condition that u⃗\vec{u} is orthogonal to (5,2,6)(5,2,6) gives:

(2α+β)5+(α−β)2+(α+β)6=18α+9β=0.(2\alpha+\beta)5 + (\alpha-\beta)2 + (\alpha+\beta)6 = 18\alpha+9\beta=0.

This simplifies to:

2α+β=0⟹β=−2α.2\alpha+\beta=0 \quad\Longrightarrow\quad \beta=-2\alpha.

Substitute β=−2α\beta=-2\alpha into u⃗\vec{u}:

u⃗=(2α−2α,  α−(−2α),  α+(−2α))=(0,  3α,  −α).\vec{u} = (2\alpha-2\alpha,\; \alpha-(-2\alpha),\; \alpha+(-2\alpha)) = (0,\; 3\alpha,\; -\alpha).

To be a unit vector, choose ∣α∣|\alpha| such that:

∥u⃗∥=02+(3α)2+(−α)2=9α2+α2=10α2=∣α∣10=1.\|\vec{u}\| = \sqrt{0^2+(3\alpha)^2+(-\alpha)^2} = \sqrt{9\alpha^2+\alpha^2} = \sqrt{10\alpha^2} = |\alpha|\sqrt{10}=1.

Thus, ∣α∣=110|\alpha|=\frac{1}{\sqrt{10}}. Choosing α=110\alpha=\frac{1}{\sqrt{10}} gives:

u⃗=(0,310,−110).\vec{u}=\left(0,\frac{3}{\sqrt{10}},\frac{-1}{\sqrt{10}}\right).

Multiplying by −1-1 (which is acceptable as the unit vector direction is unique up to sign) yields:

u⃗=(0,−310,110)=−3j^+k^10.\vec{u}=\left(0,\frac{-3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right)=\frac{-3\hat{j}+\hat{k}}{\sqrt{10}}.