MAH CET MCA Rectangular Cartesian Coordinates Previous Year Questions (PYQs)
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The perpendicular distance of the point $P(1,2,3)$ from the line $\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}$ is
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Solution
Line passes through $A(6,7,7)$ with direction $\vec v=\langle3,2,-2\rangle$.
$\vec{AP}=\langle-5,-5,-4\rangle$. Distance $=\dfrac{\lVert \vec{AP}\times\vec v\rVert}{\lVert\vec v\rVert}=\dfrac{\sqrt{833}}{\sqrt{17}}=\sqrt{49}=7$.
The projections of a line segment on the $X,Y,Z$ axes are $12,4,3$ respectively. The length and direction cosines are
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Solution
Components are $(12,4,3)$. Length $L=\sqrt{12^2+4^2+3^2}=13$.
Direction cosines $=\left(\dfrac{12}{13},\dfrac{4}{13},\dfrac{3}{13}\right)$.
If $A(cos\alpha, sin\alpha)$, $B(sin\alpha, -cos\alpha)$, C(1,2) are the vertices of a $\Delta ABC$, then as $\alpha$ varies, the the locus of its centroid is,
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The points (a: b) , (c, d) and $\frac{kc+la}{k+l}$,$\frac{kd+lb}{k+l}$ are
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Solution
MAH CET MCA
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