Question
Show that the points whose position vectors are

$\vec{\text{a}}=4\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}, \vec{\text{b}}=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}-\hat{\text{j}}$ from a right triangle.

Answer

Given
$\vec{\text{a}}=4\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}-\hat{\text{j}}$
$\overrightarrow{\text{AB}}$ = position vector of B - position vector of A
$=\big(2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)-\big(4\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}\big)$
$=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}-4\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
$\overrightarrow{\text{AB}}=-2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$
$\overrightarrow{\text{BC}}$ = position vector of C - position vector of B
$=\big(\hat{\text{i}}-\hat{\text{j}}\big)-\big(2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)$
$=\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$=-\hat{\text{i}}+3\hat{\text{j}}-5\hat{\text{k}}$
$\overrightarrow{\text{CA}}$ = position vector of A - position vector of C
$=\big(4\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)$
$=4\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}-\hat{\text{i}}+\hat{\text{j}}$
$=3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$

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