CUET UG Mathematics Applied Vector Previous Year Questions (PYQs) – Page 1 of 2
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If $\vec a,\ \vec b$ and $\vec c$ are three vectors such that $\vec a+\vec b+\vec c=\vec 0$, where $\vec a$ and $\vec b$ are unit vectors and $|\vec c|=2$, then the angle between the vectors $\vec b$ and $\vec c$ is:
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3
4
Solution
If $\cos\alpha,\cos\beta,\cos\gamma$ are the direction cosines of vector $\vec a$, then value of
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma$
is equal to:
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2
3
4
Solution
For direction cosines
$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
Using identity
$\cos 2\theta = 2\cos^2\theta -1$
So
$\cos 2\alpha + \cos 2\beta + \cos 2\gamma$
$=(2\cos^2\alpha-1)+(2\cos^2\beta-1)+(2\cos^2\gamma-1)$
$=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3$
$=2(1)-3$
$=-1$
The value of$\mathbf{i}\cdot(\mathbf{j}\times\mathbf{k})+\mathbf{j}\cdot(\mathbf{i}\times\mathbf{k})+\mathbf{k}\cdot(\mathbf{i}\times\mathbf{j})$is:
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Solution
Using vector identities:
$\mathbf{i}\times\mathbf{j}=\mathbf{k}$
$\mathbf{j}\times\mathbf{k}=\mathbf{i}$
$\mathbf{k}\times\mathbf{i}=\mathbf{j}$
First term
$\mathbf{i}\cdot(\mathbf{j}\times\mathbf{k})$
$=\mathbf{i}\cdot\mathbf{i}=1$
Second term
$\mathbf{j}\cdot(\mathbf{i}\times\mathbf{k})$
But
$\mathbf{i}\times\mathbf{k}=-\mathbf{j}$
So
$\mathbf{j}\cdot(-\mathbf{j})=-1$
Third term
$\mathbf{k}\cdot(\mathbf{i}\times\mathbf{j})$
$=\mathbf{k}\cdot\mathbf{k}=1$
Adding
$1-1+1=1$
The area of the parallelogram whose adjacent sides are
$(\mathbf{i}+\mathbf{k})$ and $(2\mathbf{i}+\mathbf{j}+\mathbf{k})$ is:
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Solution
Area of parallelogram
$= |\vec a \times \vec b|$
Let
$\vec a = (1,0,1)$
$\vec b = (2,1,1)$
Cross product
$\vec a \times \vec b =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 0 & 1 \\
2 & 1 & 1
\end{vmatrix}$
$= \mathbf{i}(0\cdot1-1\cdot1)
-\mathbf{j}(1\cdot1-1\cdot2)
+\mathbf{k}(1\cdot1-0\cdot2)$
$= -\mathbf{i}+\mathbf{j}+\mathbf{k}$
Magnitude
$|\vec a \times \vec b|
=\sqrt{(-1)^2+1^2+1^2}$
$=\sqrt{3}$
If $x(\mathbf{i}+\mathbf{j}+\mathbf{k})$ is a unit vector then value of $x$ is:
1
2
3
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Solution
Vector
$\vec v = x(\mathbf{i}+\mathbf{j}+\mathbf{k})$
Magnitude of unit vector
$|\vec v|=1$
$|x(\mathbf{i}+\mathbf{j}+\mathbf{k})|
=|x|\sqrt{1^2+1^2+1^2}$
$=|x|\sqrt{3}$
Since magnitude is 1
$|x|\sqrt{3}=1$
$|x|=\frac{1}{\sqrt{3}}$
Therefore
$x=\pm\frac{1}{\sqrt{3}}$
The equation of plane passing through the point $(0,7,-7)$ and containing the line
$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$
is:
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Solution
The line
$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$
gives a point on the line
$(-1,3,-2)$
Direction vector of line
$\vec d=(-3,2,1)$
Vector joining given point and line point
$(0,7,-7)-(-1,3,-2)=(1,4,-5)$
Normal vector of plane
$\vec n = \vec d \times (1,4,-5)$
$\vec n=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-3 & 2 & 1 \\
1 & 4 & -5
\end{vmatrix}$
$= -14\mathbf{i}-14\mathbf{j}-14\mathbf{k}$
Normal vector $\propto (1,1,1)$
Plane equation
$x+y+z+d=0$
Substitute point $(0,7,-7)$
$0+7-7+d=0$
$d=0$
Plane
$x+y+z=0$
A line
$\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-1}{-1}$
is perpendicular to a plane $(P)$ which passes through the point $(4,3,9)$.
If the mirror image of point $S$ on the line $(L)$ in the given plane $(P)$ is $(2,3,1)$, then coordinates of point $S$ is:
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Solution
Direction ratios of line
$\vec n=(1,2,-1)$
Since the line is perpendicular to the plane, this vector is the **normal vector of the plane**.
Plane passing through $(4,3,9)$
Equation
$1(x-4)+2(y-3)-1(z-9)=0$
$x+2y-z-1=0$
Mirror image property:
The plane is the midpoint between $S$ and its image.
Image point
$S'=(2,3,1)$
Parametric form of line
$x=2+t$
$y=3+2t$
$z=1-t$
So any point on the line
$S=(2+t,3+2t,1-t)$
Midpoint between $S$ and $S'$
$\left(\frac{2+t+2}{2},\frac{3+2t+3}{2},\frac{1-t+1}{2}\right)$
$=\left(2+\frac{t}{2},3+t,1-\frac{t}{2}\right)$
Substitute into plane
$x+2y-z-1=0$
$\left(2+\frac{t}{2}\right)+2(3+t)-\left(1-\frac{t}{2}\right)-1=0$
$6+3t=0$
$t=-2$
Hence
$S=(2-2,3-4,1+2)$
$S=(0,-1,3)$
A ball is thrown upwards from the plane surface of the ground.
Suppose the plane surface from which the ball is thrown consists of the points
$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.
The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.
Using this information answer the question.
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Solution
Points
$A(1,0,2)$
$B(3,-1,1)$
$C(1,2,1)$
Direction vectors
$\vec{AB}=(3-1,-1-0,1-2)=(2,-1,-1)$
$\vec{AC}=(1-1,2-0,1-2)=(0,2,-1)$
Normal vector of plane
$\vec{n}=\vec{AB}\times\vec{AC}$
$\vec{n}=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
2 & -1 & -1\\
0 & 2 & -1
\end{vmatrix}$
$=3\mathbf{i}+2\mathbf{j}+4\mathbf{k}$
So normal vector
$(3,2,4)$
Equation of plane
$3(x-1)+2(y-0)+4(z-2)=0$
$3x+2y+4z-11=0$
$3x+2y+4z=11$
A ball is thrown upwards from the plane surface of the ground.
Suppose the plane surface from which the ball is thrown consists of the points
$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.
The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.
Using this information answer the question.
The maximum height of the ball from the ground is:
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Solution
Equation of plane through $A,B,C$
$3x+2y+4z=11$
Maximum height of the ball = perpendicular distance of point $D(2,3,1)$ from the plane.
Distance formula
$d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$
Plane
$3x+2y+4z-11=0$
Substitute $D(2,3,1)$
$=|3(2)+2(3)+4(1)-11|$
$=|6+6+4-11|$
$=|5|$
Denominator
$\sqrt{3^2+2^2+4^2}$
$=\sqrt{9+4+16}$
$=\sqrt{29}$
Therefore
$d=\frac{5}{\sqrt{29}}$
A ball is thrown upwards from the plane surface of the ground.
Suppose the plane surface from which the ball is thrown consists of the points
$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.
The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.
Using this information answer the question.
The equation of the perpendicular line drawn from the maximum height of the ball to the ground is:
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3
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Solution
Ground plane passes through points
$A(1,0,2),\; B(3,-1,1),\; C(1,2,1)$
Direction vectors
$\vec{AB}=(2,-1,-1)$
$\vec{AC}=(0,2,-1)$
Normal vector of plane
$\vec{n}=\vec{AB}\times\vec{AC}$
$\vec{n}=
\begin{vmatrix}
\mathbf{i}&\mathbf{j}&\mathbf{k}\\
2&-1&-1\\
0&2&-1
\end{vmatrix}
$
$=3\mathbf{i}+2\mathbf{j}+4\mathbf{k}$
So direction ratios of perpendicular line
$(3,2,4)$
The perpendicular line passes through point
$D(2,3,1)$
Hence equation of line
$\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$
CUET UG Mathematics Applied
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CUET UG Mathematics Applied
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and More.
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