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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
For a triangle $ABC$, let $\vec{p} = \overrightarrow{BC},; \vec{q} = \overrightarrow{CA}$ and $\vec{r} = \overrightarrow{BA}$. If $|\vec{p}| = 2\sqrt{3},; |\vec{q}| = 2$ and $\cos\theta = \frac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then
$|\vec{p}\times(\vec{q}-3\vec{r})|^2 + 3|\vec{r}|^2$ is equal to:
Solution
$\vec{p} + \vec{q} = \vec{r}$
$\cos(\pi - \theta) = \frac{|\vec{p}|^2 + |\vec{q}|^2 - |\vec{r}|^2}{2|\vec{p}||\vec{q}|}$
$-\frac{1}{\sqrt{3}} = \frac{12 + 4 - |\vec{r}|^2}{2\cdot 2\sqrt{3}\cdot 2}$
$\Rightarrow |\vec{r}|^2 = 24$
$|\vec{p}\times(\vec{q}-3\vec{r})|^2 + 3|\vec{r}|^2$
$= |\vec{p}\times(\vec{q}-3(\vec{p}+\vec{q}))|^2 + 72$
$= |\vec{p}\times(-3\vec{p}-2\vec{q})|^2 + 72$
$= | -2\vec{p}\times\vec{q} |^2 + 72$
$= 4|\vec{p}|^2 |\vec{q}|^2 \sin^2\theta + 72$
$= 4\cdot 12 \cdot 4 \cdot \frac{2}{3} + 72 = 200$
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