AMU MCA Vector Previous Year Questions (PYQs) – Page 3 of 3
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If
$U={(x,y)\in\mathbb R^2\mid y=mx}$
and
$W={(x,y)\in\mathbb R^2\mid y=tx,\ t\ne m}$
Then $\dim(U+W)$ is
1
2
3
4
Solution
Both are distinct lines through origin.
Their sum spans $\mathbb R^2$.
So dimension = 2.
Let $\mathbb{R}^3={(x,y,z)}$ and $W$ be the subspace generated by ${(1,2,-3)}$. Geometrically, $W$ represents
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2
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4
Solution
Single vector span ⇒ line through origin
Direction ratios: $1,2,-3$
Direction cosines:
$\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}}$
The value of $\lambda$ so that the vectors
$a = 2\hat{i} - \hat{j} + \hat{k}$, $b = \hat{i} + 2\hat{j} - 3\hat{k}$ and
$c = 3\hat{i} + \lambda \hat{j} + 5\hat{k}$ are coplanar is
1
2
3
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Solution
Coplanar ⇒ scalar triple product = 0
$\left|\matrix{2 & -1 & 1 \cr 1 & 2 & -3 \cr 3 & \lambda & 5}\right| = 0$
Expand:
$= 2(10+3\lambda) + 1(5+9) + 1(\lambda-6)$
$= 20 + 6\lambda + 14 + \lambda - 6$
$= 28 + 7\lambda = 0$
$\lambda = -4$
If origin is the centroid of a $\triangle PQR$ with vertices
$P(2a,2,6)$, $Q(-4,3b,-10)$ and $R(8,14,2c)$, then the value of $a,b,c$ are respectively
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2
3
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Solution
Centroid = $(0,0,0)$
So:
$\frac{2a-4+8}{3}=0 \Rightarrow 2a+4=0 \Rightarrow a=-2$
$\frac{2+3b+14}{3}=0 \Rightarrow 3b+16=0 \Rightarrow b=-\frac{16}{3}$
$\frac{6-10+2c}{3}=0 \Rightarrow 2c-4=0 \Rightarrow c=2$
The angle between the lines with direction ratios $(4,-3,5)$ and $(3,4,5)$ is
1
2
3
4
Solution
$\cos\theta = \frac{4\cdot3 + (-3)\cdot4 + 5\cdot5}{\sqrt{50}\sqrt{50}} = \frac{25}{50} = \frac{1}{2}$
$\theta = \frac{\pi}{3}$
Final Answer: $\boxed{\frac{\pi}{3}}$
A particle of mass 2 kg is moving such that its position is given by
$P(t)=5\hat{i}-2t^2\hat{j}$
Find angular momentum at $t=2$ sec about origin.
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Solution
Position at $t=2$:
$\vec{r}=5\hat{i}-8\hat{j}$
Velocity:
$\vec{v}=\frac{d\vec{r}}{dt}=-4t\hat{j}$
At $t=2$:
$\vec{v}=-8\hat{j}$
Momentum:
$\vec{p}=m\vec{v}=2(-8\hat{j})=-16\hat{j}$
Angular momentum:
$\vec{L}=\vec{r}\times\vec{p}$
$=(5\hat{i}-8\hat{j})\times(-16\hat{j})$
$= -80\hat{k}$
If $A \cap B = {0}$, $\dim A = 2$, $\dim B = 1$ in 3D space, then
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Solution
$\dim(A + B) = 2 + 1 - 0 = 3$
So $V = A + B$
Let $V$ be a 3-dimensional vector space with $A$ and $B$ its subspaces of dimensions $2$ and $1$ respectively. If $A \cap B = {0}$, then
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Solution
$\dim(A + B) = 2 + 1 - 0 = 3$
So $V = A + B$
The direction cosines of the line which is equally inclined to the axes is
1
2
3
4
Solution
If equally inclined then $l = m = n$
$l^2 + m^2 + n^2 = 1$
$3l^2 = 1$
$l = \pm\frac{1}{\sqrt{3}}$
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