M.C.Q (1 Marks)

Question 1011 Mark

ox, oy are positive x-axis, positive y-axis respectively where O = (0, 0,0) The d.c.s of the llne which bisects $\angle\text{xoy}$ are:

$1,1,0$
$\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$
$\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}$
$0,0,1$

Answer

$\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$

Solution:
Equation of line bisecting XOY is x = y
$\therefore$ d.r.s are (1, 1, 0)
And thus d.c.s are $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$

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Question 1021 Mark

What is the sum of the squares ofdirection cosines of the line joining thepoints (1, 2, -3) and (-2, 3, 1):

0
1
3
$\frac{2}{\sqrt{26}}$

Answer

Solution:
The sum of the squares of direction cosines of the line is always 1

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Question 1031 Mark

If $\alpha,\beta$ and $\gamma$ are the angles which a half ray makes with the positive direction of the axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to:

1
2
0
-1

Answer

Solution:
Given expression, $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$
$=(1-\cos^2\alpha)+(1-\cos^2\beta)+(1-\cos^2\gamma)$
$=3-\cos^2\alpha+\cos^2\beta+\cos^2\gamma=3-1=2$
$(\because\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1)$

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Question 1041 Mark

The direction ratios of the line x - y + z - 5 = 0 = x - 3y - 6 are proportional to:

$3,1,-2$
$2,-4,1$
$\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{-2}{\sqrt{14}}$
$\frac{2}{\sqrt{41}},\frac{-4}{\sqrt{41}},\frac{1}{\sqrt{41}}$

Answer

$3,1,-2$

Solution:
We have
x - y + z - 5 = 0 = x - 3y - 6
⇒ x - 3y - 6=0
x - y + z - 5 = 0
⇒ x = 3y + 6 .....(1)
x - y + z - 5 = 0.....(2)
From (1) and (2), we get
3y + 6 - y + z - 5 = 0
⇒ 2y + z + 1 = 0
$\Rightarrow\text{y}=\frac{-\text{z}-1}{2}$
$\text{y}=\frac{\text{x}-6}{3}$ [From (1)]
$\therefore\frac{\text{x}-6}{3}=\text{y}=\frac{-\text{z}-1}{2}$
So, the given equation can be re-witten as
$\frac{\text{x}-6}{3}=\frac{\text{y}}{1}=\frac{\text{z}+1}{-2}$
Hence, the direction ratios the given line are proportional to 3, 1, -2.

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Question 1051 Mark

If $\alpha,\beta,\gamma$ are the angles which a directed line makes with the positive directions of the coordinate axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ is equal to:

Answer

Solution
The direction cosines of the line are
$\text{l}^2+\text{m}^2+\text{n}^2=1$
Now, $\Rightarrow\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$
$\Rightarrow1-\sin^2\alpha+1-\sin^2\beta+1-\sin^2\gamma=1$
$\Rightarrow\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$

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Question 1061 Mark

If a line makes angle $\frac{\pi}{3}$ and $\frac{\pi}{4}$ with x-axis and y-axis respectively, then the angle made by the line with z-axis is:

$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{5\pi}{12}$

Answer

$\frac{\pi}{3}$

Solution:
If a line makes angles $\alpha,\beta$ and $\gamma$ with the axcs, then $\cos2\alpha+\cos2\beta+\cos2\gamma=1.$
Here,
$\alpha=\frac{\pi}{3}$
$\beta=\frac{\pi}{4}$
Now,
$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$
$\Rightarrow\cos^2\frac{\pi}{3}+\cos^2\frac{\pi}{4}+\cos^2\gamma=1$
$\Rightarrow\frac{1}{4}+\frac{1}{2}+\cos^2\gamma=1 $
$\Rightarrow\cos^2\gamma=1-\frac{3}{4}$
$\Rightarrow\cos^2\gamma=\frac{1}{4}$
$\Rightarrow\cos\gamma=\frac{1}{2}$
$\Rightarrow\gamma=\frac{\pi}{3}$

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Question 1071 Mark

The shortest distance between the lines $\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}$ and, $\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}$ is:

$\sqrt{30}$
$2\sqrt{30}$
$5\sqrt{30}$
$3\sqrt{30}$

Answer

$3\sqrt{30}$

Solution:
We have
$\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}\dots(1)$
$\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}\dots(2)$
We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, -1, 1.
Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1,$ where $\vec{\text{a}}_1=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}_1=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$
Also, line (2) passes through the point (3, -7, 6) and has direction ratios proprtional to -3, 2, 4.
Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ where $\vec{\text{a}}_2=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}.$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-1&1\\-3&2&4\end{vmatrix}$
$=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-6)^2+(-15)^2+3^2}$
$=\sqrt{36+225+9}$
$=\sqrt{270}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big).\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big)$
$=36+225+9$
$=270$
The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{270}{\sqrt{270}}\Big|$
$=\sqrt{270}$
$=3\sqrt{30}$

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Question 1081 Mark

The projection of a directed line segment on the co-ordinate axes are 12, 4, 3, then the direction cosines of the line are:

$\frac{-12}{13},\frac{-4}{13},\frac{-3}{13}$
$\frac{12}{13},\frac{4}{13},\frac{3}{13}$
$\frac{12}{13},\frac{-4}{13},\frac{3}{13}$
$\frac{12}{13},\frac{4}{13},\frac{-3}{13}$

Answer

$\frac{12}{13},\frac{4}{13},\frac{3}{13}$

Solution:
x = 12, y = 4, z = 3
Direction cosines =
$\frac{\text{x}}{\text{x}^2 + \text{y}^2+\text{z}^2},\frac{\text{y}}{\text{x}^2 + \text{y}^2+\text{z}^2},\frac{\text{x}}{\text{x}^2 + \text{y}^2+\text{z}^2}$
$=\frac{12}{13},\frac{4}{13},\frac{3}{13}$

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Question 1091 Mark

(2, -3, -1) 2x - 3y + 6z + 7 = 0:

4
3
2
$\frac{1}{5}$

Answer

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MCQ 1101 Mark

If the projections of the line segment $AB$ on the coordinate axes are $12, 3, k$ and $AB = 13$ then $k^2 - 2k + 3$ is equal to$:$

A
$0$
B
$1$
C
$11$
✓
$27$

Answer

Correct option: D.

$27$

Let $a, b, c$ be the projection of a line on the coordinate axes.
Then the length of the line given by $\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}$
Here we have $12^2 + 3^2 + k^2 = 169$
$\Rightarrow\text{k}=\underline{+}4$
Thus $k^2 - 2k + 3 = 11$ or $27.$

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MCQ 1111 Mark

The distance between the planes $2x + 2y - z +2 = 0$ and $4x + 4y - 2z + 5 = 0$ is:

A
$\frac{1}{2}$
B
$\frac{1}{4}$
✓
$\frac{1}{6}$
D
$\text{None of these}$

Answer

Correct option: C.

$\frac{1}{6}$

Multiplying the first equation of the plane by
$4x + 4y - 2z + 4 = 0$
$4x + 4y - 2z = -4 .....(1)$
The second eqution of the plane is
$4x + 4y - 2z + 5 = 0$
$4x + 4y - 2z = -5 .....(2)$
We know that the distance between two planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is,
$=\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|-5+4|}{\sqrt{4^2+4^2+(-2)^2}}$
$=\frac{|-1|}{\sqrt{16+16+4}}$
$=\frac{1}{\sqrt{36}}$
$=\frac{1}{6}\text{units}$

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Question 1121 Mark

The direction cosines of the y-axis are:

(9, 0, 0)
(1, 0, 0)
(0, 1, 0)
(0, 0, 1)

Answer

(0, 1, 0)

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Question 1131 Mark

lf a line makes angles $\frac{\pi}{12},\frac{5\pi}{12}$ with oy, oz respectively where 0 = (0, 0, 0), then the angle made by that line with ox is:

45°
90°
60°
30°

Answer

90°

Solution:
$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{5\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1$
$\Big(\cos\frac{\pi}{12}\Big)^2+\Big(\cos\frac{\pi}{12}\Big)^2+\big(\cos(\gamma)\big)^2=1..$
$\Big(\cos\theta=\sin\Big(\frac{\pi}{2}- \theta\Big)\Big)$
$\Big(\cos(\gamma)\Big)^2=0$
$\cos(\gamma)=0$
$\gamma=90^\circ$

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MCQ 1141 Mark

The direction ratios of the line joining the points $(x, y, z)$ and $(x_2, y_2, z_1)$ are$:$

A
$\text{x}_{1} + \text{x}_{2}, \text{y}_{1} +\text{ y}_{2}, \text{z}_{1} + \text{z}_{2}$
B
$ (\text{x}_{1}-\text{x}_{2})^2+(\text{y}_{1}-\text{y}_{2})^2+(\text{z}_{1}+\text{z}_{2})^2$
C
$\frac{\text{x}_{1}+\text{x}_{2}}{2}, \frac{\text{y}_{1}+\text{y}_{2}}{2}, \frac{\text{z}_{1}+\text{z}_{2}}{2}$
✓
$\text{x}_{2} - \text{x}_{1}, \text{y}_{2} - \text{y}_{1}, \text{z}_{2} -\text{ z}_{1}$

Answer

Correct option: D.

$\text{x}_{2} - \text{x}_{1}, \text{y}_{2} - \text{y}_{1}, \text{z}_{2} -\text{ z}_{1}$

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Question 1151 Mark

If a line makes the angle $\alpha,\beta,\gamma$ with three dimensional coordinate axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma$ is equal to:

-2
-1
1
2

Answer

-1

Solution:
We need to find value of $\cos2\alpha+\cos2\beta+\cos2\gamma$
It is further equal to $\cos^2\alpha-1+\cos^2\beta-1+\cos^2\gamma-1$
$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3$
$= 2(1) - 3 = 2 = -1$
$\therefore(\text{l}^2 + \text{m}^2 + \text{n}^2 = 1)$

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Question 1161 Mark

The projection of the join of the two points (1, 4, 5), (6, 7, 2) on the line whose d.ss are (4, 5, 6) is:

$\frac{17}{\sqrt{77}}$
$\frac{7}{6}$
$21$
$\frac{7}{9}$

Answer

$\frac{17}{\sqrt{77}}$

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Question 1171 Mark

lf $\text{AB}\perp\text{BC}$ then the value of $\lambda$ equal, where A(2k, 2, 3), B(k, 1, 5), C(3 + k, 2, 1):

$3$
$\frac{1}{3}$
$-3$
$-\frac{1}{3}$

Answer

$-3$

Solution:
The drs of AB are (k, 1, -2)
The drs of BC are (3, 1, -4)
Since, they are perpendicular, AB.BC = 0
3k + 1 + 8 = 0
k = -3

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Question 1181 Mark

The direction ratios of the line perprndicular to the lines $\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$ and, $\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$ are proportional to:

4, 5, 7
4, -5, 7
4, -5, -7
-4, 5, 7

Answer

4, 5, 7

Solution:
We have
$\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$
$\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$
The direction ratios of the given lines are proportional to 2, -3, 1 and 1, 2, -2.
The vectors parallel to the given vectors are $\vec{\text{b}}_1=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
Vector perpendicular to the given two lines is
$\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&1\\1&2&-2\end{vmatrix}$
$=4\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$
Hence, the direction ration of the line perpendicular to the given two lines are proportional to 4, 5, 7.

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Question 1191 Mark

Choose the correct answer
Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is:

$2\ \text{units}$
$4\ \text{units}$
$8\ \text{units}$
$\frac{2}{\sqrt{29}}\ \text{units}.$

Answer

Equation of one plane is 2x + 3y + 4z = 4
⇒ 2x + 3y + 4z - 4 = 0
Equation of second plane is 4x + 6y + 8z = 12
⇒ 4x + 6y + 8z - 12 = 0
⇒ 2x + 3y + 4z - 6 = 0
Here $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2},\ \frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2},\ \frac{\text{c}_1}{\text{c}_2}=\frac{4}{8}=\frac{1}{2}$
Since, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ therefore, the given two lines are parallel.
We know that the distance of parallel lines $=\frac{|\text{d}_1-\text{d}_2|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
$\Rightarrow\ \frac{|-4-(-6)|}{\sqrt{(2)^2+(3)^2+(4)^2}}$
$=\frac{|-4+6|}{\sqrt{4+9+16}}$
$=\frac{2}{\sqrt{29}}$
Therefore, option (D) is correct.

$\frac{2}{\sqrt{29}}.$

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MCQ 1201 Mark

Choose the correct answer The planes: $2x – y + 4z = 5$ and $5x – 2.5y + 10z = 6$ are:

A
Perpendicular
✓
Parallel
C
Intersect $y-$axis
D
Passes through $\Big(0,\ 0,\ \frac{5}{4}\Big).$

Answer

Correct option: B.

Parallel

Equation of the given planes are $2x - y + 4z = 5(a_1x + b_1y + c_1z + d = 0)$
and $5x - 2.5y + 10z = 6(a_2x + b_2y + c_2z + d = 0)$
For perpendicular $a_1a_2 + b_1b_2 + c_1c_2 $
$= 2(5) + (-1)(-2.5) + 4(10) $
$= 10 + 2.5 + 40 $
$= 52.5$
$\because\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2\neq0$
$\therefore$ Planes are not perpendicular.
For parallel $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{5},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-1}{-2.5}=\frac{10}{25}=\frac{2}{5},\ \frac{\text{c}_1}{\text{c}_2}=\frac{4}{10}=\frac{2}{5}$
$\because\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore$ given planes are parallel.
Therefore, option $(B)$ is correct.

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Question 1211 Mark

The direction cosines of the ray P(1, -2, 4) and Q(-1, 1, -2) are:

$\big(-2, -3, -6\big)$
$\big(2, -3, -6\big)$
$\Big(\frac{2}{7},\frac{3}{7},\frac{6}{7}\Big)$
$\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$

Answer

$\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$

Solution:
P(1, -2, 4), Q(-1, 1, -2)
$\text{PQ}=\sqrt{(1-(1))^2 +(2-1)^2+(4-(-2))^{2}}$
$=\sqrt{4+9+36}$
$=\sqrt{49}=7\text{DC}$
$=\Big(\frac{-1-1}{7},\frac{1-(2)}{7},\frac{-2-4}{7}\Big)$
$=\Big(-\frac{2}{7},\frac{3}{7},-\frac{6}{7}\Big)$

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Question 1221 Mark

The length of the perpendicular drawn from the point (4, -7, 3) on the y-axis is:

3 units
4 units
5 units
7 units

Answer

5 units

Solution:
The length of the perpendicular drawn from the point (4, -7, 3) on the y-axis is
⇒ Point on the y-axis would be = (0, -7, 0)
The length of the perpendicular drawn $=\sqrt{(4-0)^2}+(-7-(-7))^2+(3-0)^2$
$=\sqrt{4^2}+0^2+3^2$
$\Rightarrow\sqrt{16}+0+9$
$=\sqrt{25}$
$=5$

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Question 1231 Mark

The points with position vectors 60i + 3j, 40i − 8j and ai − 52j are collinear if:

a = −40
a = 40
a = 20
None of these

Answer

a = −40

Solution:
Denoting a,b,c by the given vectors respectively
These vectors will be collinear if there is some constant k such that c − a = K(b − a)
⇒ a − 60 = −20K and −55 = −11K
⇒ a = −100 + 60 = −40

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Question 1241 Mark

Cosine of the angle between two diagonals of acube is equal to:

$\frac{2}{\sqrt{6}}$
$\frac{1}{3}$
$\frac{1}{2}$
None of these

Answer

$\frac{1}{3}$

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Question 1251 Mark

If a line makes angles $\alpha,\beta,\gamma,\delta$ with four diagonals of a cube, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta$ is equal to:

$\frac{1}{3}$
$\frac{2}{3}$
$\frac{4}{3}$
$\frac{8}{3}$

Answer

$\frac{4}{3}$

Solution:

Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure. Clearly, OP, AR
The direction ratiosm of OP, AR, BS and CQ are
a - 0, a - 0, a - 0, i.e. a, a, a
0 - a, a - 0, a - 0, i.e. -a, a, a
a - 0, 0 - a, a - 0, i.e. a, -a, a
a - 0, a - 0, 0 - a, i.e. a, a, -a
Let the direction ratios of a line be proportional to l, m and n. Suppose this line makes angles $\alpha,\beta,\gamma$ and $\delta$ with OP, AR.
Now, $\alpha$ is the angle between OP and the line whose direction ratios are proportional to l, m and n.
$\cos\alpha=\frac{\text{a}.\text{l}+\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\alpha=\frac{\text{l}+\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
Since $\beta$ is the angle between AR and the line with direction ratios proportional to l, m and n, we get
$\cos\beta=\frac{-\text{a}.\text{l}+\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\beta=\frac{-\text{l}+\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
Similarly,
$\cos\gamma=\frac{\text{a}.\text{l}-\text{a}.\text{m}+\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\gamma=\frac{\text{l}-\text{m}+\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\cos\delta=\frac{\text{a}.\text{l}+\text{a}.\text{m}-\text{a}.\text{n}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}\Rightarrow\cos\delta=\frac{\text{l}+\text{m}-\text{n}}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+\cos^2\delta$
$=\frac{(\text{l}+\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(-\text{l}+\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(\text{l}-\text{m}+\text{n})^2}{3(\text{l}^2+\text{m}^2+\text{n}^2)}+\frac{(\text{l}+\text{m}-\text{n})^2}{\sqrt{3}\sqrt{\text{l}^2+\text{m}^2+\text{n}^2}}$
$=\frac{1}{3(\text{l}^2+\text{m}^2+\text{n}^2)}\Big\{(\text{l}+\text{m}+\text{n})^2+(-\text{l}+\text{m}+\text{n})^2+(\text{l}-\text{m}+\text{n})^2+(\text{l}+\text{m}-\text{n})^2\Big\}$
$=\frac{1}{3(\text{l}^2+\text{m}^2+\text{n}^2)}4\big(\text{l}^2+\text{m}^2+\text{n}^2\big)=\frac{4}{3}$.

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Question 1261 Mark

The coordinates of the midpoints of the line segment joining the points (2, 3, 4) and (8, -3, 8) are:

(10, 0, 12)
(5, 6, 0)
(6, 5, 0)
(5, 0, 6)

Answer

(5, 0, 6)

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Question 1271 Mark

The distance of the plane through the intersection of the planes ax + by + cz +d = 0 and lx + my + nz + P = 0 and parallel to the line y = 0, z = 0

(bl - am)y + (cl - an)z + dl - ap = 0
(am - bl)x + (mc - bn)z + md - bp = 0
(na - cl)x + (bn - cm)y + nd - cp = 0
None of these

Answer

(bl - am)y + (cl - an)z + dl - ap = 0

Solution:
The equation of the plane passing through the intersection of the planes
ax + by + cz + d = 0
and lx + my + nz + p =0
Will be $(\text{ax} + \text{by} +\text{cz} +\text{d})+\lambda(\text{lm}+\text{my}+\text{nz}+\text{p})=0$
$\text{x}(\text{a}+\lambda1)+\text{y}(\text{b}+\lambda\text{m})+\text{z}(\text{c}+\lambda\text{n})+(\text{d}+\lambda\text{p})=0\ (1)$
Since the plane is parallel to the line y = 0 and z = 0
$\text{a}+\lambda1=0$
$\lambda=\frac{-\text{a}}{\text{l}}$
Putting the value of A in eqution (1), we get
$\text{x}\Big(\text{a}+\Big(\frac{\text{-a}}{\text{l}}\Big)\text{l}\Big)+\text{y}\Big(\text{b}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{m}+\text{y}\Big(\text{c}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{n}+\text{d}+\Big(\frac{-\text{a}}{\text{l}}\Big)\text{p}=0$
$\text{y}(\text{bl}-\text{am})+\text{z}(\text{cl}-\text{an})+\text{dl}-\text{ap}=0$
Heance, option (a)

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Question 1281 Mark

The eqution of the plane contaning the two lines $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}}{-2}=\frac{\text{y}-2}{-3}=\frac{\text{z}+1}{-1}$ is:

8x + y - 5z - 7 = 0
8x + y + 5z - 7 = 0
8x - y - 5z - 7 = 0
None of these

Answer

None of these

Solution:
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-0}{3}$ and $\frac{\text{x}}{-2}=\frac{\text{y}-2}{-3}=\frac{\text{z}+1}{-1}$
Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.

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Question 1291 Mark

The perpendicular distance of the point P(1, 2, 3) from the line $\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$ is:

7
5
0
None of these

Answer

Solution:
$\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$
Let point (1, 2, 3) be P and the point through which the line passes be Q(6, 7, 7). Also, the line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
Now,
$\overrightarrow{\text{PQ}}=5\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}} =\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&2&-2\\5&5&4\end{vmatrix}$
$=18\hat{\text{i}}-22\hat{\text{j}}+5\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{18^2+(-22)^2+5^2}$
$=\sqrt{324+484+25}$
$=\sqrt{833}$
$\because\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{833}}{\sqrt{17}}$
$=\sqrt{49}$
$=7$

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Question 1301 Mark

The area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2) is equal to:

9sq. units
1sq. units
27sq. units
81sq. units

Answer

9sq. units

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Question 1311 Mark

How many lines through the origin in make equal angles with the coordinate axis:

Answer

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MCQ 1321 Mark

The equation $x^2- x - 2 = 0$ in three dimensional space is represented by$:$

✓
A pair of parallel planes
B
A pair of straight lines
C
A pair of perpendicular plane
D
None of these

Answer

Correct option: A.

A pair of parallel planes

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Question 1331 Mark

A straight line L on the xy-plane bisects the angle between OX and OY. What are the direction cosines of L:

$\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$
$\Big(\frac{1}{2},\frac{\sqrt{3}}{2},0\Big)$
$\big(0,0,1\big)$
$\Big(\frac{2}{3},\frac{2}{3},\frac{1}{3}\Big)$

Answer

$\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$

Solution
L makes an angle $\frac{\pi}{4}$ with X and Y axis and $\frac{\pi}{2}$
$\therefore$ d.cs are $\Big(\cos\frac{\pi}{34},\cos\frac{\pi}{4},\cos\frac{\pi}{2}\Big)=\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$

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Question 1341 Mark

A line passes through the points (6, −7, −1) and (2, −3, 1). The direction cosines of the line so directed that the angle made by it with the positive direction of x-axis is acute, is?

$\frac{2}{3},\frac{2}{3},-\frac{1}{3}$
$-\frac{2}{3},\frac{2}{3},\frac{1}{3}$
$\frac{2}{3}-\frac{2}{3},\frac{1}{3}$
$\frac{2}{3},\frac{2}{3},\frac{1}{3}$

Answer

$\frac{2}{3},\frac{2}{3},-\frac{1}{3}$

Solution:
Consider the problem
Let l, m, n are direction cosines of the given line.
then as it made an acute angle with x−axis,
Therefore, l > 0
The line passes through (6, −7, −1) and (2, −3, 1)
Therefore, its direction ratios are
6 − 2, −7 + 3, −1−1 or 2, −2, −1
Hence direction cosines of the line are given by $\frac{2}{3},\frac{2}{3},-\frac{1}{3}.$

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Question 1351 Mark

If the directions cosines of a line are A, k, k, then:

k > 0
0 < k < 1
k = 1
$\text{k}=\frac{\sqrt{1}}{3}$ or $\frac{\sqrt{1}}{3}$

Answer

$\text{k}=\frac{\sqrt{1}}{3}$ or $\frac{\sqrt{1}}{3}$

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Question 1361 Mark

The length of the $\perp$

0
$2\sqrt{3}$
$\frac{2}{3}$
2

Answer

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Question 1371 Mark

The sine of the angle between the straight line $\frac{\text{x}-2}{3}=\frac{\text{y}-3}{4}=\frac{\text{z}-4}{5}$ and the plane 2x - 2y + z = 5 is:

$\frac{10}{6\sqrt{5}}$
$\frac{4}{5\sqrt{2}}$
$\frac{2\sqrt{3}}{5}$
$\sqrt{\frac{\sqrt{2}}{10}}$

Answer

$\frac{2\sqrt{3}}{5}$

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Question 1381 Mark

A line with positive direction cosines passes through the point P(2, -1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals:

$1$
$\sqrt{2}$
$\sqrt{3}$
$2$

Answer

$\sqrt{3}$

Solution:
D.C of the line are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
Any point on the line at a distance tt from P(2, -1, 2) is
$\Big(2+\frac{\text{t}}{\sqrt{3}},-1+\frac{\text{t}}{\sqrt{3}},2+\frac{\text{t}}{\sqrt{3}}\Big)$
which lies on $2\text{x} + \text{y + z} = 9$
$\Rightarrow\text{t}=\sqrt{3}$

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Question 1391 Mark

The direction cosines of the line joining (1, -1, 1) and (-1, 1, 1) are:

2, -2, 0
1, -1, 0
$\frac{1}{\sqrt{2}},- \frac{1}{\sqrt{2}}$
None of these

Answer

$\frac{1}{\sqrt{2}},- \frac{1}{\sqrt{2}}$

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Question 1401 Mark

If the projections of the line segment AB on the coordinate axes are 2, 3, 6, then the square of the sine of the angle made by AB with x = 0, is:

$\frac{3}{7}$
$\frac{3}{49}$
$\frac{4}{7}$
$\frac{40}{49}$

Answer

$\frac{40}{49}$

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Question 1411 Mark

What are the direction cosines of a line which is equally inclined to the positive directions of the axes:

$\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$
$\Big(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$
$\Big(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\Big)$
$\Big(\frac{1}{3},\frac{1}{3},\frac{1}{3}\Big)$

Answer

$\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$

Solution:
We know sum of the squares of the direction cosines is one.
i.e. $\cos^2\alpha+\cos^2\gamma=1$
but its given that $\alpha=\beta=\gamma\therefore\cos^2\alpha=1$
$3\cos^2\alpha=1$
$\therefore\cos^2\alpha=\frac{1}{3}$
$\therefore$ Positive directions of the axes are $\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big)$

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Question 1421 Mark

Direction cosines of ray from P(1, -2, 4) to Q(-1, 1, -2) are:

-2, 3, -6
2, -3, 6
2, 3, 6
$\frac{-2}{7},\frac{3}{7},\frac{-6}{7}$

Answer

$\frac{-2}{7},\frac{3}{7},\frac{-6}{7}$

Solution:
Given the points are P(1, -2, 4) and Q(-1, 1, -2) Now the direction
ratios of the ray PQ are (-1 - 1, 1 + 2, -2 - 4) = (-2, 3, -6)
The direction cosines of the line PQ will be
$\Big(\frac{2}{\sqrt{2^2+3^2+6^2}},\frac{3}{\sqrt{2^2+3^2+6^2}},\frac{-6}{\sqrt{2^2+3^2+6^2}}\Big)=\Big(\frac{-2}{7},\frac{3}{7},\frac{-6}{7}\Big)$

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Question 1431 Mark

The plane $2\text{x}-(1-\lambda)\text{y}+3\lambda\text{z}=0$ passes through the intersection of the planes:

2x - y = 0 and y- 3z = 0
2x + 3z = 0 and y = 0
2x - y + 3z = 0 and y - 3z = 0
None of these

Answer

2x - y = 0 and y- 3z = 0

Solution:
The given plane is
$2\text{x}-(1-\lambda)\text{y}+3\lambda\text{z}=0$
$\Rightarrow(2\text{x}-\text{y})+\lambda(-\text{y}+3\text{z})=0$
So, this plane passes through the intersection of the planes
2x - y = 0 and -y + 3z = 0
⇒ 2x - y = 0 and y - 3z = 0.

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Question 1441 Mark

What are the direction ratios of the line if it passes through the intersection of the planes x = 3z + 4 and y = 2z - 3:

(1, 2, 3)
(2, 1, 3)
(3, 2, 1)
(1, 3, 2)

Answer

(3, 2, 1)

Solution:
Equations of the planes are x = 3z + 4 and y = 2z - 3
$\therefore$ The equation of the plane passing through the line of intersection of these planes is x = 3z + 4 and y = 2z - 3
Thus The direction Ratios of the equation passes through intersection of the planes is (3, 2, 1).

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Question 1451 Mark

The product of the d.cs of the line which makesequal angles with ox, oy, oz is:

$1$
$\sqrt{3}$
$\frac{1}{3\sqrt{3}}$
$\frac{1}{\sqrt{3}}$

Answer

$\frac{1}{3\sqrt{3}}$

Solution:
$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=1$
$\Rightarrow3\cos^2(\alpha)=1$
$\Rightarrow\cos\alpha=\underline{+}\frac{1}{\sqrt{3}}$

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Question 1461 Mark

The direction cosines of the straight linegiven by the planes x = 0 and z = 0 are:

1, 0, 0
0, 0, 1
1, 1, 0
0, 1, 0

Answer

0, 1, 0

Solution:
Given, x = z = 0
It represents Z-axis
$\therefore$ Direction cosines = (0, 1, 0)

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Question 1471 Mark

Choose the correct answer from the given four options.
The sine of the angle between the straight line $\frac{\text{x}-2}{3}=\frac{\text{y}-2}{3}=\frac{\text{z}-2}{3}$ and the plane $2\text{x}-2\text{y}+\text{z}=5$ is:

$\frac{10}{6\sqrt{5}}$
$\frac{4}{5\sqrt{2}}$
$\frac{2\sqrt{3}}{5}$
$\frac{\sqrt{2}}{10}$

Answer

$\frac{\sqrt{2}}{10}$

Solution:
We have, the equation of line as
$\frac{\text{x}-2}{3}=\frac{\text{y}-2}{3}=\frac{\text{z}-2}{3}$
This line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$
Equation of plane is $2\text{x}-2\text{y}+\text{z}=5$
Normal to the plane $\vec{\text{n}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
Its angle between line and plane is $'\theta'.$
Then $\sin\theta=\frac{|\vec{\text{b}}\cdot{\vec{\text{b}}}|}{|{\vec{\text{b}}}||{\vec{\text{b}}}|}$
$=\frac{\big|(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}})\cdot(2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}})\big|}{\sqrt{3^2+4^2+5^2}\sqrt{4+4+1}}$
$=\frac{|6-8+5|}{\sqrt{50}\sqrt{9}}$
$=\frac{3}{15\sqrt{2}}=\frac{1}{5\sqrt{2}}$
$\sin\theta=\frac{\sqrt{2}}{10}$

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Question 1481 Mark

The length of perpendicular from the origin to the plane which makes intercepts $\frac{1}{3},\frac{1}{4},\frac{1}{5}$ respectively on the coordinate axes is:

$\frac{1}{\sqrt[5]{2}}$
$\frac{1}{10}$
$\sqrt[5]{2}$
$5$

Answer

$\frac{1}{\sqrt[5]{2}}$

Solution:
Equation of plane $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
3x + 4y + 5z − 1 = 0
diatance from origin $\frac{1}{\sqrt{150}}=\frac{1}{\sqrt[5]{2}}$

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Question 1491 Mark

The straigth line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:

parallel to x-axis
parallel to y-axis
parallel to z-axis
perpendicular to z-axis

Answer

perpendicular to z-axis

Solution:
We have
$\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$
Also, the given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be parpendicular to the given line.
Now,
$3\text{x}+4\text{y}+0\text{z}=0$
It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).
Hence, the given line is perpendicular to z-axis.

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Question 1501 Mark

For every point P(x, y, z) on the xy-plane,

x = 0
y = 0
z = 0
x = y = z = 0

Answer

z = 0

Solution:
The Z-coordinate of every point on the XY-plane is zero.

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MATHS M.C.Q (1 Marks)