First slide

Direction ratios and direction cosines

Question

If the vertices of a triangle are $A (1, 2, 1), B (4, 3, 1), C (3, 1, 5)$ then $\frac{Area of Δ A B C}{42 \times \cos^{2} A} =$

Moderate

A

1
B

5
C

10
D

15

Solution

The vertices of a triangle are $A (1, 2, 1), B (4, 3, 1), C (3, 1, 5)$
The area of the triangle   $A B C$ is $\frac{1}{2} |\bar{A B} \times \bar{A C}|$
Here $\bar{A B} = 3 i + j and \bar{A C} = 2 i - j + 4 k$
Hence the area of the triangle is
$\begin{matrix} \frac{1}{2} |\begin{matrix} i & j & k \\ 3 & 1 & 0 \\ 2 & - 1 & 4 \end{matrix}| = \frac{1}{2} \{i (4) - j (12) + k (- 3 - 2)\} \\ = \frac{1}{2} (4 i - 12 j - 5 k) \end{matrix}$
Therefore, the area of the triangle is $\frac{\sqrt{16 + 144 + 25}}{2} = \frac{\sqrt{185}}{2}$
And
$\begin{matrix} \cos A = \frac{\bar{A B} \cdot \bar{A C}}{|\bar{A B}| |\bar{A C}|} \\ = \frac{6 - 1}{\sqrt{9 + 1} \sqrt{4 + 1 + 16}} \\ = \frac{5}{\sqrt{10} \sqrt{21}} \\ = \frac{\sqrt{5}}{\sqrt{42}} \end{matrix}$
Consider
$\frac{Area of Δ A B C}{42 \times \cos^{2} A} = \frac{\sqrt{185}}{2 \times 42 \times \frac{5}{42}} = \frac{\sqrt{185}}{10}$

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