AMU MCA Vector Previous Year Questions (PYQs) – Page 2 of 3
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Showing 29 questions
Which statement is true for the line
$\frac{x - 4}{8} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and plane having intercepts $-4, 2$ and $3$ of the following
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Solution
The image of the point $ (1, -1, 1) $ in the plane
$ x - 2y + 3z + 1 = 0 $ is
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Solution
The length of perpendicular from the point $\hat{i}+2\hat{j}+3\hat{k}$ to the line
$\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ is
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Solution
The unit vector which is orthogonal to the vector $\hat{i}+\hat{j}+\hat{k}$ and is coplanar with vectors $\hat{i}+2\hat{j}-\hat{k}$ and $2\hat{i}+\hat{j}+3\hat{k}$ is
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The distance of point of intersection of the line
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ with the plane $x-y+z=5$ from the point with position vector $\hat{i}-2\hat{j}+3\hat{k}$ is
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The angle between the line
$x-2y+z=0=x+2y-2z$ and the plane $5x-2y-z+17=0$ is
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Solution
The shortest distance between the lines
$\vec r=(4\hat i-\hat j)+\lambda(\hat i+2\hat j-3\hat k)$
and
$\vec r=(\hat i-\hat j+2\hat k)+\mu(2\hat i+4\hat j-5\hat k)$ is
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Solution
Direction vectors:
$ \vec b_1=(1,2,-3),\ \vec b_2=(2,4,-5) $
$ \vec b_1\times \vec b_2=(2,-1,0) $
$ |\vec b_1\times \vec b_2|=\sqrt{4+1}=\sqrt{5} $
Points:
$ \vec a_1=(4,-1,0),\ \vec a_2=(1,-1,2) $
$ \vec a_2-\vec a_1=(-3,0,2) $
Shortest distance:
$ d=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times \vec b_2)|}{|\vec b_1\times \vec b_2|} $
$ =\frac{|(-3,0,2)\cdot(2,-1,0)|}{\sqrt{5}} $
$ =\frac{|-6|}{\sqrt{5}}=\frac{6}{\sqrt{5}} $
If the sum of two unit vectors is again a unit vector, then magnitude of their difference is
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Solution
$ |\vec a+\vec b|=1,\ |\vec a|=|\vec b|=1 $
$ |\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b $
$ 1=1+1+2\vec a\cdot\vec b $
$ \Rightarrow \vec a\cdot\vec b=-\frac{1}{2} $
Now,
$ |\vec a-\vec b|^2=1+1-2\left(-\frac{1}{2}\right)=3 $
$ \Rightarrow |\vec a-\vec b|=\sqrt{3} $
Let $A$ and $B$ be two points with position vectors $\vec a$ and $\vec b$ respectively and let $C$ be a point dividing $AB$ internally and the position vector of $C$ on $AB$ is $\vec c=\lambda \vec a+\mu \vec b$, then
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Solution
Using section formula,
$ \vec c=\frac{m\vec b+1\cdot \vec a}{m+1} $
Comparing with $\vec c=\lambda \vec a+\mu \vec b$,
$ \lambda=\frac{1}{m+1},\ \mu=\frac{m}{m+1} $
$ \Rightarrow \lambda+\mu=\frac{1+m}{m+1}=1 $
Which of the following is a 2-dimensional subspace of $\mathbb{R}^3$?
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Solution
Set ${(0,x,z)}$ contains two independent parameters and satisfies subspace properties.
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