The cosine of the angle $A$ of the triangle with vertices $A(1,-1,2), B(6,11,2), C(1,2,6)$ is
The cosine of the angle $A$ of the triangle with vertices $A(1,-1,2), B(6,11,2), C(1,2,6)$ is
- $63 / 65$
- $36 / 65$
- $16 / 65$
- $13 / 64$
Solution
Direction ratios of
$\begin{aligned} A B & =6-1,11+1,2-2 \\ & =5,12,0\end{aligned}$
Direction ratios of
$\begin{aligned} A C & =1-1,2+1,6-2 \\ & =0,3,4\end{aligned}$
Now, $\begin{aligned} \cos A & =\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\ \Rightarrow \quad \cos A & =\frac{5 \times 0+12 \times 3+0 \times 4}{\sqrt{25+144+0} \sqrt{0+9+16}} \\ & =\frac{36}{13 \times 5} \\ & =\frac{36}{65}\end{aligned}$
Asked in: AP EAMCET 2007
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