Question

The volume of parallelopiped, whose coterminous edges are given by $\overline{u}=\hat{i}+\hat{j}+\lambda\hat{k}, \vec{v}=\hat{i}+\hat{j}+3\hat{k}, \overline{w}=2\hat{i}+\hat{j}+\hat{k}$ is 1 cu. units. If $\theta$ is the angle between $\overline{u}$ and $\overline{w}$, then the value of $\cos\theta$ is

• A. $\frac{3}{4}$
• B. $\frac{5}{6}$
• C. $\frac{1}{5}$
• D. $\frac{1}{6}$

Hint:

Logic Tip: The problem yields two possible values for $\lambda$ ($\lambda=2$ or $\lambda=4$). If $\lambda=4$, $\cos\theta = (2+1+4)/(\sqrt{18}\sqrt{6}) = 7/\sqrt{108} = 7/(6\sqrt{3})$, which isn't in the options. In multiple-choice questions, proceed immediately with the simplest integer root first; it usually maps to the correct option.

Answer 1

The Correct Option is B

Concept:
The volume of a parallelepiped formed by vectors $\vec{u}, \vec{v}, \vec{w}$ is: \[ |[\vec{u}\ \vec{v}\ \vec{w}]| \] The angle between two vectors is given by: \[ \cos\theta = \frac{\vec{u} \cdot \vec{w}}{|\vec{u}|\,|\vec{w}|} \] 
Step 1: Find $\lambda$ using volume.
\[ V = \begin{vmatrix} 1 & 1 & \lambda \\ 1 & 1 & 3 \\ 2 & 1 & 1 \end{vmatrix} \] \[ V = 1(1\cdot1 - 3\cdot1) - 1(1\cdot1 - 3\cdot2) + \lambda(1\cdot1 - 1\cdot2) \] \[ V = 1(1 - 3) - 1(1 - 6) + \lambda(1 - 2) \] \[ V = -2 + 5 - \lambda = 3 - \lambda \] Given $|V| = 1$: \[ |3 - \lambda| = 1 \Rightarrow \lambda = 2 \text{ or } 4 \] Take $\lambda = 2$. 
Step 2: Form vectors and find magnitudes.
\[ \vec{u} = \hat{i} + \hat{j} + 2\hat{k}, \quad \vec{w} = 2\hat{i} + \hat{j} + \hat{k} \] \[ |\vec{u}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}, \quad |\vec{w}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6} \] 
Step 3: Compute $\cos\theta$.
\[ \vec{u} \cdot \vec{w} = 1\cdot2 + 1\cdot1 + 2\cdot1 = 5 \] \[ \cos\theta = \frac{5}{\sqrt{6}\cdot\sqrt{6}} = \frac{5}{6} \] 
Final Answer:
\[ \boxed{\frac{5}{6}} \]