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Vector Algebra Part 2

Section: Mathematics

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Question : 8 of 100

Marks: +1, -0

Let for a triangle

ABC

\;\overrightarrow{AB} = -2\hat{i}+\hat{j}+3\hat{k}

\;\overrightarrow{CB} = \alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}

\;\overrightarrow{CA} = 4\hat{i}+3\hat{j}+\delta\hat{k}

\delta>0

and the area of the triangle

ABC

5\sqrt{6}

\overrightarrow{CB}\cdot\overrightarrow{CA}

[13-Apr-2023 shift 2]

108
60
54
120

Solution:

\;\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}

\;\langle 4,3,\delta\rangle \cdot +\langle -2,1,3\rangle=\overrightarrow{CB}

\;\Rightarrow \overrightarrow{CB}=\langle 2,4,3+\delta\rangle

\;\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-2&1&3\\-4&-3&-\delta\end{vmatrix}

\;=\hat{\ell}(9-\delta)-\hat{j}(2\delta+12)+k(10)

\;|AB\times AC|^2=(9-\delta)^2+(2\delta+12)^2+(10)^2

\;=5\delta^2+30\delta+325

\;\;\text{Area of }\triangle ABC=5\sqrt{6}

\;\Rightarrow \;\frac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}|=5\sqrt{6}

\;\Rightarrow |\overrightarrow{AB}\times\overrightarrow{AC}|^2=600

\;\Rightarrow 5\delta^2+30\delta-275=0

\;\Rightarrow S^2+6\delta-55=0

\;\Rightarrow (\delta+11)(\delta-5)=0

\;\delta=5

\;\overrightarrow{CB}=\langle 2,3,8\rangle

\;\overrightarrow{CB}\cdot\overrightarrow{CA}=\langle 2,4,8\rangle\cdot\langle 4,3,5\rangle

\;=8+12+40=60

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