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: Expand
Let \(\vec u = \vec a\times\vec b\). Then \(\vec u\times(\vec c\times\vec d) = \vec c(\vec u\cdot\vec d) - \vec d(\vec u\cdot\vec c)\).

Step 2: Use coplanarity
The vectors are coplanar, so \(\vec u\perp\vec c\) and \(\vec u\cdot\vec c = 0\). Also \(\vec d\) is perpendicular to the plane, so \(\vec d\) is parallel to \(\vec u\).

Step 3: Magnitude
\(|\vec u| = \sin30^{\circ} = \frac12\), so \(\vec u\cdot\vec d = \pm\frac12\). The expression becomes \(\pm\frac12\vec c\).

Step 4: Solve
\(\vec c = 2\left(\frac{3}{26}\hat i-\frac2{13}\hat j+\frac6{13}\hat k\right) = \frac3{13}\hat i-\frac4{13}\hat j+\frac{12}{13}\hat k\) (taking the positive sign).

Step 5: Check
Its length: \(\frac{9+16+144}{169} = 1\), so it is a unit vector. Option (A). The other options are not unit vectors.

Final Answer:
c = (3/13) i - (4/13) j + (12/13) k. \[ \boxed{\text{(A)}\ \frac3{13}\hat i-\frac4{13}\hat j+\frac{12}{13}\hat k} \]