Let for a triangle ABC, A B = - 2 i ^ + j ^ + 3 k ^ , C B = i ^ + j ^ + k ^ , C A = 4 i ^ + 3 j ^ + k ^ . If 0 and the area of the triangle ABC is 5 6 , then C B · C A is equal to [2023]

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Solution

Q.
Let for a triangle ABC, $\vec{A B} = - 2 \hat{i} + \hat{j} + 3 \hat{k},$ $\vec{C B} = α \hat{i} + β \hat{j} + γ \hat{k},$ $\vec{C A} = 4 \hat{i} + 3 \hat{j} + δ \hat{k} .$ If $δ > 0$ and the area of the triangle ABC is $5 \sqrt{6}$ , then $\vec{C B} \cdot \vec{C A}$ is equal to [2023]

1 120

2 60

3 108

4 54

Ans.
(2)

Given, $\vec{A B} = - 2 \hat{i} + \hat{j} + 3 \hat{k}$

$\vec{C B} = α \hat{i} + β \hat{j} + γ \hat{k}$

$\vec{C A} = 4 \hat{i} + 3 \hat{j} + δ \hat{k}$

$In triangle, \vec{A B} + \vec{B C} + \vec{C A} = \vec{0}$

$- 2 \hat{i} + \hat{j} + 3 \hat{k} - α \hat{i} - β \hat{j} - γ \hat{k} + 4 \hat{i} + 3 \hat{j} + δ \hat{k} = 0$

$\Rightarrow \hat{i} (2 - α) + \hat{j} (4 - β) + \hat{k} (3 - γ + δ) = 0$

$∴$ $α = 2, β = 4, γ - δ = 3$

$Area of triangle =$ $| \frac{1}{2} | \vec{A B} \times \vec{A C} | |$ $= 5 \sqrt{6}$

$\Rightarrow$ $\frac{1}{2}$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 1 & 3 \\ - 4 & - 3 & - δ \end{matrix} |$ $= 5 \sqrt{6}$

$\Rightarrow$ ${(δ - 9)}^{2} + {(2 δ + 12)}^{2} + 100 = 600$

$\Rightarrow δ = 5 and γ = 8$

Hence, $\vec{C B} \cdot \vec{C A} = (2 \hat{i} + 4 \hat{j} + 8 \hat{k}) \cdot (4 \hat{i} + 3 \hat{j} + 5 \hat{k}) = 60$

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