Question

Let \(\overset{̄}{a},\overset{̄}{b}\) and \(\overset{̄}{c}\) be three coplanar unit vectors. A unit vector \(\overset{̄}{d}\) is perpendicular to them. If \((\overset{̄}{a}\times \overset{̄}{b})\times (\overset{̄}{c}\times \overset{̄}{d}) = \frac{3}{26}\hat{i}-\frac{2}{13}\hat{j}+\frac{6}{13}\hat{k}\) and the angle between \(\overset{̄}{a}\) and \(\overset{̄}{b}\) is \(30^{\circ}\), then \(\overset{̄}{c}\) is equal to...

• A. \(\frac{3}{13}\hat{i}-\frac{4}{13}\hat{j}+\frac{12}{13}\hat{k}\)
• B. \(\frac{3}{13}\hat{i}-\frac{2}{13}\hat{j}+\frac{6}{13}\hat{k}\)
• C. \(\frac{3}{26}\hat{i}-\frac{4}{13}\hat{j}+\frac{12}{13}\hat{k}\)
• D. \(\frac{3}{26}\hat{i}-\frac{3}{26}\hat{j}+\frac{5}{26}\hat{k}\)

Hint:

Expand u cross (c cross d) with u = a cross b and use that c lies in the plane of a and b.

Answer 1

The Correct Option is A

Step 1: Which options are unit vectors
Compute squared lengths: (A) $\frac{9+16+144}{169} = 1$. (B) $\frac{9+4+36}{169} = \frac{49}{169}$. (C) $\frac{9}{676}+\frac{16}{169}+\frac{144}{169}$ is not 1. (D) $\frac{9+9+25}{676}$ is not 1.

Step 2: Link with the expansion
The identity reduces the left side to $\pm\frac12\vec c$, so $\vec c$ must be a unit vector equal to twice the right side up to sign.

Step 3: Result
Twice the given vector is option (A), a unit vector.

Final Answer:
Option A. \[ \boxed{\text{(A)}\ \frac3{13}\hat i-\frac4{13}\hat j+\frac{12}{13}\hat k} \]