If the vertices of a triangle areA1,2,1,B4,3,1,C3,1,5 then  Area of Δ ABC42×cos2A=

# If the vertices of a triangle are$A\left(1,2,1\right),B\left(4,3,1\right),C\left(3,1,5\right)$ then  $\frac{\text{Area\hspace{0.17em}of\hspace{0.17em}}\Delta \text{\hspace{0.17em}}ABC}{42×{\mathrm{cos}}^{2}A}=$

1. A

1

2. B

5

3. C

10

4. D

15

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### Solution:

The vertices of a triangle are$A\left(1,2,1\right),B\left(4,3,1\right),C\left(3,1,5\right)$

The area of the triangle  $ABC$is$\frac{1}{2}\left|\overline{AB}×\overline{AC}\right|$

Here$\overline{AB}=3i+j\text{\hspace{0.17em}and }\overline{AC}=2i-j+4k$

Hence the area of the triangle is

$\begin{array}{c}\frac{1}{2}\left|\begin{array}{ccc}i& j& k\\ 3& 1& 0\\ 2& -1& 4\end{array}\right|=\frac{1}{2}\left\{i\left(4\right)-j\left(12\right)+k\left(-3-2\right)\right\}\\ =\frac{1}{2}\left(4i-12j-5k\right)\end{array}$

Therefore, the area of the triangle is  $\frac{\sqrt{16+144+25}}{2}=\frac{\sqrt{185}}{2}$

And

$\begin{array}{c}\mathrm{cos}A=\frac{\overline{AB}\cdot \overline{AC}}{\left|\overline{AB}\right|\left|\overline{AC}\right|}\\ =\frac{6-1}{\sqrt{9+1}\sqrt{4+1+16}}\\ =\frac{5}{\sqrt{10}\sqrt{21}}\\ =\frac{\sqrt{5}}{\sqrt{42}}\end{array}$

Consider

$\frac{\text{Area\hspace{0.17em}of\hspace{0.17em}}\Delta \text{\hspace{0.17em}}ABC}{42×{\mathrm{cos}}^{2}A}=\frac{\sqrt{185}}{2×42×\frac{5}{42}}=\frac{\sqrt{185}}{10}$

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