Question
Find $\big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\big]$, when
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{k}}$

Answer

Given:
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}$
$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{c}}=3\hat{\text{i}}-\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=(2\hat{\text{i}}-3\hat{\text{j}})\times(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})$
$=2\hat{\text{k}}+2\hat{\text{j}}+3\hat{\text{k}}+3\hat{\text{i}}$
$=3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}=\big(3\hat{\text{i}}+2\hat{\text{j}}+5\hat{\text{k}}\big).\big(3\hat{\text{i}}-\hat{\text{k}}\big)$
$=9-5=4\ ....(1)$
Now,
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}$
$=4$ [Using (1)]

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