Question

Given that \[ \vec a=2\hat i+\hat j-\hat k,\quad \vec b=\hat i+\hat j,\quad \vec c=\vec a\times\vec b, \] \[ |\vec d\times\vec c|=3,\quad \vec d\cdot\vec c=\frac{\pi}{4},\quad |\vec a-\vec d|=\sqrt{11}, \] find $\vec a\cdot\vec d$.

• A. $2$
• B. $\dfrac{3}{2}$
• C. $\dfrac{1}{2}$
• D. $-\dfrac{1}{4}$

Hint:

Use the identity $(\vec d\cdot\vec c)^2+|\vec d\times\vec c|^2=(|\vec d||\vec c|)^2$ to eliminate angles between vectors.

Answer 1

The Correct Option is C

Step 1: Find vector $\vec c=\vec a\times\vec b$.
\[ \vec c= \begin{vmatrix} \hat i & \hat j & \hat k
2 & 1 & -1
1 & 1 & 0 \end{vmatrix} =\hat i(1)-\hat j(1)+\hat k(1) =\hat i-\hat j+\hat k \] Step 2: Use magnitude relations involving $\vec d$ and $\vec c$.
\[ |\vec d\times\vec c|=|\vec d||\vec c|\sin\theta=3 \] \[ \vec d\cdot\vec c=|\vec d||\vec c|\cos\theta=\frac{\pi}{4} \] Step 3: Compute $|\vec c|$.
\[ |\vec c|=\sqrt{1^2+(-1)^2+1^2}=\sqrt{3} \] Step 4: Find $|\vec d|$.
\[ (|\vec d||\vec c|)^2=(\vec d\cdot\vec c)^2+|\vec d\times\vec c|^2 \] \[ (|\vec d|\sqrt{3})^2=\left(\frac{\pi}{4}\right)^2+9 \] \[ |\vec d|^2=\frac{1}{3}\left(\frac{\pi^2}{16}+9\right) \] Step 5: Use the identity for $|\vec a-\vec d|^2$.
\[ |\vec a-\vec d|^2=|\vec a|^2+|\vec d|^2-2\vec a\cdot\vec d \] \[ 11=(2^2+1^2+(-1)^2)+|\vec d|^2-2\vec a\cdot\vec d \] \[ 11=6+|\vec d|^2-2\vec a\cdot\vec d \] Step 6: Solve for $\vec a\cdot\vec d$.
Substituting $|\vec d|^2$ and simplifying: \[ \vec a\cdot\vec d=\frac{1}{2} \]