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If the vertices $A, B, C$ of a triangle $ABC$ are $(1, 2, 3), (-1, 0, 0), (0, 1, 2)$ respectively, then find $ \angle ABC. \angle ABC$ is the angle between the vectors $ \overline {BA}$ and \( \overline {BC}\) ].

$\begin{array}{1 1}(A) \cos^{-1}\big(\large\frac{10}{\sqrt{102}}\big) \\ (B) \cos^{-1}\big(\large\frac{-10}{\sqrt{102}}\big) \\(C) \cos^{-1}\big(\large\frac{10}{102}\big) \\ (D) \cos^{-1}\big(\large\frac{-10}{102}\big)\end{array} $

1 Answer

  • $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|\cos\theta$
  • $\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Step 1:
Let O be the origin,then position vector
$\overrightarrow {OA}=\hat i+2\hat j+3\hat k$
$\overrightarrow {OB}=-\hat i$
$\overrightarrow {OC}=\hat j+2\hat k$
Step 2:
$\overrightarrow {BC}=\overrightarrow {OC}-\overrightarrow{OB}$
$\quad\;\;=(\hat j+2\hat k)-(-\hat i)$
$\quad\;\;=\hat i+\hat j+2\hat k$
Step 3:
$\overrightarrow {BA}=\overrightarrow {OA}-\overrightarrow{OB}$
$\quad\;\;=(\hat i+2\hat j+3\hat k)-(-\hat i)$
$\quad\;\;=2\hat i+2\hat j+3\hat k$
Step 4:
Now let us find $\angle ABC$
We know $\cos\theta=\large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Here $\cos\angle ABC=\large\frac{\overrightarrow{BC}.\overrightarrow{BA}}{|\overrightarrow{BC}||\overrightarrow{BA}|}$
$\qquad\qquad\qquad=\large\frac{(\hat i+\hat j+2\hat k)(2\hat i+2\hat j+3\hat k)}{|\hat i+\hat j+2\hat k||2\hat i+2\hat j+3\hat k|}$
$\qquad\qquad\qquad=\large\frac{1\times 2+1\times 2+2\times 3}{\sqrt{1^2+1^2+2^2}\sqrt{2^2+2^2+3^2}}$
$\qquad\qquad\qquad=\large\frac{2+2+6}{\sqrt 6\sqrt{17}}$
Step 5:
$\cos\angle ABC=\large\frac{10}{\sqrt{102}}$
$\angle ABC=\cos^{-1}\big(\large\frac{10}{\sqrt{102}}\big)$
answered May 21, 2013 by sreemathi.v

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