Question

Let for a triangle ABC,
\(\vec{AB}=-2\hat{i}+\hat{j}+3\hat{k} \)
\(\vec{CB}=α\hat{i}+β\hat{j}+γ\hat{k} \)
\(\vec{CA}=4\hat{i}+3\hat{j}+δ\hat{k} \)
If \(δ>0\) and the area of the triangle ABC is \(5\sqrt6\), then \(\vec{CB}.\vec{CA}\) is equal to

• A. 60
• B. 54
• C. 120
• D. 108

Answer 1

The Correct Option is A

$\overrightarrow{CB} = \overrightarrow{AB} + \overrightarrow{CA} = (-2+4)\hat{i} + (1+3)\hat{j} + (3+\delta)\hat{k} = 2\hat{i} + 4\hat{j} + (3+\delta)\hat{k}$ \item $\overrightarrow{CB} \times \overrightarrow{CA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 2 & 4 & 3+\delta 4 & 3 & \delta \end{vmatrix} = (\delta-9)\hat{i} + (12+2\delta)\hat{j} - 10\hat{k}$ 

 $|\overrightarrow{CB} \times \overrightarrow{CA}| = \sqrt{(\delta-9)^2 + (12+2\delta)^2 + 100}$ 

Area = $\frac{1}{2}|\overrightarrow{CB} \times \overrightarrow{CA}| = 5\sqrt{6}$

$|\overrightarrow{CB} \times \overrightarrow{CA}| = 10\sqrt{6}$ 

$\sqrt{(\delta-9)^2 + (12+2\delta)^2 + 100} = 10\sqrt{6}$ 

$(\delta-9)^2 + (12+2\delta)^2 + 100 = 600$ 

$\delta^2 - 18\delta + 81 + 4\delta^2 + 48\delta + 144 + 100 = 600$ 

$5\delta^2 + 30\delta + 325 = 600$ \item $5\delta^2 + 30\delta - 275 = 0$ 

$\delta^2 + 6\delta - 55 = 0$ 

$(\delta+11)(\delta-5) = 0$ 

$\delta = 5$ (since $\delta > 0$) 

$\overrightarrow{CB} = 2\hat{i} + 4\hat{j} + 8\hat{k}$ 

$\overrightarrow{CB} \cdot \overrightarrow{CA} = (2)(4) + (4)(3) + (8)(5) = 8 + 12 + 40 = 60$ 

Answer: $\overrightarrow{CB} \cdot \overrightarrow{CA} = 60$