CUET UG Mathematics Applied Vector Previous Year Question Paper and Solution with PDF

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CUET UG Mathematics Applied
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CUET UG Mathematics Applied
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CUET UG Mathematics Applied

CUET UG Mathematics Applied Previous Year Questions (PYQs)

CUET UG Mathematics Applied Vector PYQ

Qus : 1 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2024 Mathematics PYQ

Solution

Qus : 2 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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 $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ $=(2\cos^2\alpha-1)+(2\cos^2\beta-1)+(2\cos^2\gamma-1)$ $=2(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)-3$ $=2(1)-3$ $=-1$

Qus : 3 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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 $1-1+1=1$

Qus : 4 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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 $|\vec a \times \vec b| =\sqrt{(-1)^2+1^2+1^2}$ $=\sqrt{3}$

Qus : 5 CUET UG Mathematics Applied PYQ

If $x(\mathbf{i}+\mathbf{j}+\mathbf{k})$ is a unit vector then value of $x$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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 $x=\pm\frac{1}{\sqrt{3}}$

Qus : 6 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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

$x+y+z=0$

Qus : 7 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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

$S=(2-2,3-4,1+2)$

$S=(0,-1,3)$

Qus : 8 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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

$3(x-1)+2(y-0)+4(z-2)=0$

$3x+2y+4z-11=0$

$3x+2y+4z=11$

Qus : 9 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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

$d=\frac{5}{\sqrt{29}}$

Qus : 10 CUET UG Mathematics Applied PYQ

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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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

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

$\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}$

Qus : 11 CUET UG Mathematics Applied PYQ

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 coordinates of the foot of the perpendicular drawn from the maximum height of the ball to the ground are:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

Plane through points $A,B,C$ $3x+2y+4z=11$ Normal vector of plane $\vec{n}=(3,2,4)$ Point $D(2,3,1)$ Foot of perpendicular formula $\left(x_1-\frac{a(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2},\; y_1-\frac{b(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2},\; z_1-\frac{c(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}\right)$ Plane $3x+2y+4z-11=0$ Substitute point $D$ $3(2)+2(3)+4(1)-11$ $=6+6+4-11$ $=5$ Denominator $3^2+2^2+4^2$ $=9+4+16$ $=29$ Coordinates of foot $x=2-\frac{3(5)}{29}=\frac{43}{29}$ $y=3-\frac{2(5)}{29}=\frac{77}{29}$ $z=1-\frac{4(5)}{29}=\frac{9}{29}$ Foot of perpendicular $\left(\frac{43}{29},\frac{77}{29},\frac{9}{29}\right)$

Qus : 12 CUET UG Mathematics Applied PYQ

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 area of $\triangle ABC$ is:

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CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ

Solution

$A(1,0,2)$ $B(3,-1,1)$ $C(1,2,1)$ Vectors $\vec{AB}=(3-1,-1-0,1-2)=(2,-1,-1)$ $\vec{AC}=(1-1,2-0,1-2)=(0,2,-1)$ Cross product $\vec{AB}\times\vec{AC}$ $= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&-1&-1\\ 0&2&-1 \end{vmatrix} $ $=3\mathbf{i}+2\mathbf{j}+4\mathbf{k}$ Magnitude $|\vec{AB}\times\vec{AC}|$ $=\sqrt{3^2+2^2+4^2}$ $=\sqrt{29}$ Area of triangle $=\frac{1}{2}|\vec{AB}\times\vec{AC}|$ $=\frac{1}{2}\sqrt{29}$

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