AMU MCA Vector Previous Year Questions (PYQs) – Page 1 of 3
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Showing 29 questions
The shortest distance from the point $(1,0,-2)$ to the plane
$x+2y+z=4$ is
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Solution
Distance formula:
$d=\frac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}$
Plane: $x+2y+z-4=0$
$d=\frac{|1(1)+2(0)+1(-2)-4|}{\sqrt{1^2+2^2+1^2}}$
$=\frac{|1-2-4|}{\sqrt6}=\frac{5}{\sqrt6} =\frac{5\sqrt6}{6}$
The equation of the plane through the intersection of the planes $x + y + z = 1$ and $2x + 3y - z + 4 = 0$ and parallel to $X$-axis is
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Solution
If $A, B, C, D$ are the points $(2,3,-1), (3,5,-3), (1,2,3), (3,5,7)$ respectively, then the angle between $AB$ and $CD$ is
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Solution
Two bodies of mass 100 g and 20 g are moving with velocities
$(2\hat{i}-7\hat{j}+3\hat{k})$ cm/s and $(-10\hat{i}+35\hat{j}-3\hat{k})$ cm/s respectively.
What is the velocity of centre of mass of this two body system?
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Solution
Velocity of centre of mass:
$\vec{V}=\frac{m_1\vec{v}_1+m_2\vec{v}_2}{m_1+m_2}$
$=\frac{100(2\hat{i}-7\hat{j}+3\hat{k})+20(-10\hat{i}+35\hat{j}-3\hat{k})}{120}$
$=\frac{(200-200)\hat{i}+(-700+700)\hat{j}+(300-60)\hat{k}}{120}$
$=\frac{240\hat{k}}{120}=2\hat{k}$
If $\alpha, \beta, \gamma$ be the angles which a line subtends with the positive direction of coordinate axes, then
$\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$ equals
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Solution
Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors, then $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \times \vec{c}$ if
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If $ \hat{i}, \hat{j}, \hat{k} $ are unit vectors and $ |\vec{a}| = a $, then the value of
$ |\hat{i} \times \vec{a}|^2 + |\hat{j} \times \vec{a}|^2 + |\hat{k} \times \vec{a}|^2 $ is
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Solution
If Laplace transform $L{F(x)}=\frac1{s}e^{-3s}$, then $L{e^{-x}F(3x)}$ is
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Solution
Let the unit vectors $ \vec{a} $ and $ \vec{b} $ be perpendicular to each other and the unit vector $ \vec{c} $ be inclined at an angle $ \theta $ to both $ \vec{a} $ and $ \vec{b} $. If
$ \vec{c} = \alpha \vec{a} + \beta \vec{b} + \gamma (\vec{a} \cdot \vec{b}) $, where $ \alpha, \beta, \gamma $ are scalars, then
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Solution
If the centre of mass of three particles of masses 10, 20 and 30 gram be at the point $(1,-2,3)$, where should a fourth particle of 40 gram be placed so that the combined centre of mass may be $(1,1,1)$ ?
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Solution
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