Three vertices of a triangle are $A(0,-1,-2), B(3,1,4)$ and $C(5,7,1)$. Show that it is a right - angled triangle. Also, find its other two angles.

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Three vertices of a triangle are $A(0,-1,-2), B(3,1,4)$ and $C(5,7,1)$ . Show that it is a right – angled triangle. Also, find its other two angles.

CBSE | Class 12 | Maths | RS Aggarwal | Scalar, or Dot, Product of Vectors

Three vertices of a triangle are $A(0,-1,-2), B(3,1,4)$ and $C(5,7,1)$ . Show that it is a right – angled triangle. Also, find its other two angles.

Solution:

$\overrightarrow{O A}=-\hat{\jmath}-2 \hat{k}$
$\overrightarrow{O B}=3 \hat{\imath}+\hat{\jmath}+4 \hat{k}$
$\overrightarrow{O C}=5 \hat{\imath}+7 \hat{\jmath}+\hat{k}$
$\overrightarrow{O D}=\hat{\imath}+2 \hat{\jmath}$
$\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=3 \hat{\imath}+2 \hat{\jmath}+6 \hat{k}$
$\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}=2 \hat{\imath}+6 \hat{\jmath}-3 \hat{k}$
$\overrightarrow{A C}=5 \hat{\imath}+8 \hat{\jmath}+3 \hat{k}$
$\overrightarrow{A B} \mathrm{I}=\sqrt{3^{2}+2^{2}+6^{2}}=7$
$\mid \overrightarrow{B C} I=\sqrt{2^{2}+6^{2}+3^{2}}=7$
$\overrightarrow{I A C}=\sqrt{5^{2}+8^{2}+3^{2}}=7 \sqrt{2}$
Clearly, $\overrightarrow{\mathrm{IB}} \mathrm{I}_{2}+\mathrm{I} \overrightarrow{B C} \mathrm{I}_{2}=\mathrm{I} \overrightarrow{A C} \mathrm{I}_{2}$
The given triangle is right angled triangle
$\begin{array}{l} \overrightarrow{A B} \cdot \overrightarrow{A C}=\overrightarrow{\mathrm{I}} \overrightarrow{A B} \mathrm{I} \mid \overrightarrow{A C} \mathrm{I} \cos \theta \\ 15+16+18=7.7 \sqrt{2} \cos \theta \\ \cos \theta=\frac{1}{\sqrt{2}} \\ \theta=\frac{\pi}{4} \end{array}$
Also, angle between $\overrightarrow{B C}$ and $\overrightarrow{A C}$ will be $\frac{\pi}{4}$