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MATHS M.C.Q (1 Marks)

Three Dimensional Geometry · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 180 questions.

Questions · Page 2 of 4

M.C.Q (1 Marks)

Question 511 Mark
If P(x, y, z) moves such that x = 0, z = 0 then the locus of P is the line whose d.cs are:
  1. y-axis
  2. 1, 0, 0
  3. 0, 1, 0
  4. 0, 0, 0
Answer
  1. 0, 1, 0
Solution:
When P moves then x = 0, z = 0 but y is not given. Let y = y Then the coordinates of the point will be (0, y, 0) Now, direction cosines with respect to (0, y, 0) is given by.
$\cos\alpha=\frac{0}{0^2+\text{y}^2+0^2}=\frac{0}{\text{y}}=0$
$\cos\beta=\frac{\text{y}}{0^2+\text{y}^2+0^2}=\frac{\text{y}}{\text{y}}=1$
$\cos\gamma=\frac{{0}}{0^2+\text{y}^2+0^2}=\frac{{0}}{\text{y}}=0$
The direction cosines are 0, 1, 0
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Question 521 Mark
The equation xy = 0 in three dimensional space is represented by:
  1. A plane
  2. Two plane are right angles
  3. A pair of parallel planes
  4. A pair of st. line
Answer
  1. Two plane are right angles
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Question 531 Mark
A rectangular parallelopiped is formed by planes drawn through the point (5, 7, 9) and (2, 3, 7) parallel to the coordinate planes. The length of an edge of this rectangular parallelopiped is:
  1. 2
  2. 3
  3. 4
  4. all of these
Answer
  1. all of these
Solution:
The give point (5, 7, 9) and (2, 3, 7) are two diagonally opposite vertices of the parallelopiped as all of theire coordinates.
Edges of the paralleloppiped = |5 - 2|, |7 - 3|, |9 - 7|
=3, 4, 2.
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Question 541 Mark
The distance of the line $\vec{\text{r}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}+\lambda(\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})$ from the plane $\vec{\text{r}}.(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}})=5$ is:
  1. $\frac{5}{3\sqrt{3}}$
  2. $\frac{10}{3\sqrt{3}}$
  3. $\frac{25}{3\sqrt{3}}$
  4. $\text{None of these}$
Answer
  1. $\frac{10}{3\sqrt{3}}$
Solution:
The given line passes through the point whose position vector is $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
We know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}.\vec{\text{n}}=\text{d}$ is given by
$\text{P}=\frac{\big|\vec{\text{a}}.\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
Here, $\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\vec{\text{n}}=\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}},\text{d}=5$
So, the required distance P is given by
$\text{P}=\frac{\Big|\big(2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big),\big(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\big)-5\Big|}{\Big|\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\Big|}$
$=\frac{|2-10+3-5|}{\sqrt{1+25+1}}$
$=\frac{|-10|}{\sqrt{27}}$
$=\frac{10}{3\sqrt{3}}\text{units}$
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Question 551 Mark
The eqution of the plane $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$ in scalar product from is:
  1. $\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
  2. $\vec{\text{r}}.(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=7$
  3. $\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})=7$
  4. $\text{None of these}$
Answer
  1. $\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
Solution:
We know that the equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ represents a plane passing through a point whose position vectors is $\vec{\text{a}}$ and parrallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$.
Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}};\ \vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}};\ \vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&1\\1&-2&3\end{vmatrix}$
$=5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
The vector equation of the plane in scalar product from is
$\vec{\text{r}}.\vec{\text{n}}=\vec{\text{a}}.\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(\hat{\text{i}}-\hat{\text{j}}-0\hat{\text{k}}\big).\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=5+2+0$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$
$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=7$
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Question 561 Mark
The angle between the two diagonals of a cube is:
  1. $30^\circ$
  2. $45^\circ$
  3. $\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
  4. $\cos^{-1}\Big(\frac{1}{{3}}\Big)$
Answer
  1. $\cos^{-1}\Big(\frac{1}{{3}}\Big)$
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, Consider the diagonals OP and AR.
Direction ratios of OP and AR are proportional to a - 0, a - 0, a - 0 and 0 - a, a - 0, a - 0, e.i. a, a, a and -a, a, a, respectivelly.
Let $\theta$ be the angle between OP and AR. Then,
$\cos\theta=\frac{\text{a}\times-\text{a}+\text{a}\times\text{a}+\text{a}\times\text{a}}{\sqrt{\text{a}^2+\text{a}^2+\text{a}^2}\sqrt{(-\text{a})^2+\text{a}^2+\text{a}^2}}$
$\Rightarrow\cos\theta=\frac{-\text{a}+\text{a}^2+\text{a}^2}{\sqrt{3\text{a}^2}\sqrt{3\text{a}^2}}$
$\Rightarrow\cos\theta=\frac{1}{3}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{3}\Big)$
Similarly, the angles between other pairs of the diagonals are equal to $\cos^{-1}\Big(\frac{1}{3}\Big)$ as the angle between any two diagonals.
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MCQ 571 Mark
Which of the following represents direction cosines of the line$:$
  • A
    $0,\frac{1}{\sqrt{2}},\frac{1}{2}$
  • B
    $0,\frac{-\sqrt{3}}{2},\frac{1}{\sqrt{2}}$
  • ✓
    $0,\frac{\sqrt{3}}{2},\frac{1}{2}$
  • D
    $\frac{1}{2},\frac{1}{2},\frac{1}{2}$
Answer
Correct option: C.
$0,\frac{\sqrt{3}}{2},\frac{1}{2}$
If direction cosine of a line is $l, m, n$ then
$l^2 + m^2 + n^2 = 1$
$=0^2+\Big(\frac{\sqrt{3}}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2=1$
The correct answer from the given alternative is $0,\frac{\sqrt{3}}{2},\frac{1}{2}$
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Question 581 Mark
The reflection of the point $(\text{a}, \beta, \gamma) $ in the xy-plane is:
  1. $(\alpha,\beta,0)$
  2. $(0,0,\gamma)$
  3. $(-\alpha,-\beta,\gamma)$
  4. $(\alpha,\beta,\gamma)$
Answer
  1. $(\alpha,\beta,\gamma)$
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MCQ 591 Mark
A line passes through the points $(6, -7, -1)$ and $(2, -3, 1).$ What are the direction ratios of the line?
  • A
    $(4, −4, 2)$
  • B
    $(4, 4, 2)$
  • ✓
    $(−4, 4, 2)$
  • D
    $(2, 1, 1)$
Answer
Correct option: C.
$(−4, 4, 2)$
Direction ratios of a line passing through points $(x_1​, y_1​, z_1​)$ and $(x_{2​}, y_{2​}​, z_{2​}​)$ are represented by $±(x_1​−x_{2​}​, y_{1​}−y_{2​}​, z_{1​}−z_{2​}​)$
Hence for the given line, direction ratios are $(6 − 2, −7−(−3), −1−1)$
$\Rightarrow ±(4, −4, −2)$
$\Rightarrow (−4, 4, 2)$ or $(4, −4, −2)$
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Question 601 Mark
If P be the point (2, 6, 3) then the equation of the plane trough P, at right angles to OP, where ′O′ is the origin is:
  1. 2x + 6y + 3z = 7
  2. 2x − 6y + 3z = 7
  3. 2x + 6y − 3z = 49
  4. 2x + 6y + 3z = 49
Answer
  1. 2x + 6y + 3z = 49
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Question 611 Mark
The d.rs of the lines x = ay + b, z = cy + d are:
  1. 1, a, c
  2. a, 1, c
  3. b, 1, c
  4. c, a, 1
Answer
  1. a, 1, c
Solution:
Given x = ay + b and z = cy + d
$\Rightarrow\frac{\text{x}-\text{b}}{\text{a}}=\text{y}$​ and $\frac{\text{z}-\text{d}}{\text{c}}=\text{y}$
$\Rightarrow\frac{\text{x}-\text{b}}{\text{a}}=\frac{\text{y}}{1}=\frac{\text{z}-\text{d}}{c}$
Therefore Drs of given line is a, 1, c
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Question 621 Mark
The following lines are $\hat{\text{r}}=\Big(\hat{\text{i}}+\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\Big)+\mu\Big(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\Big)$
  1. Collinear
  2. Skew-lines
  3. Co-planar lines
  4. Parallel lines
Answer
  1. Collinear
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MCQ 631 Mark
Are the points $(1, 1), (2, 3)$ and $(8, 11)$ collinear?
  • A
    collinear
  • ✓
    Non collinear
  • C
    coplaner
  • D
    None of above
Answer
Correct option: B.
Non collinear
Area of triangle formed by these vertices is,
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\2&3&1\\8&11&1\end{vmatrix}$
Applying $R_2​ \rightarrow R_2​ − R_1​, R_{3​ }\rightarrow R_{3 ​}− R1​$
$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\1&2&0\\7&10&0\end{vmatrix}=\frac{1}{2}(10-14)=2$
Hence points are non collinear
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Question 641 Mark
The direction cosines of a line which is equally inclined to axes, is given by:
  1. $\underline{+}\frac{1}{3}$
  2. $\underline{+}\frac{1}{\sqrt{3}}$
  3. $1$
  4. $0$
Answer
  1. $\underline{+}\frac{1}{\sqrt{3}}$
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Question 651 Mark
Let a vector $\vec{\text{r}}$ make angles 60°, 30° with it and y-axes respectively. Find the angle $\vec{\text{r}}$ make with z-axis:
  1. 30°
  2. 60°
  3. 90°
  4. 120°
Answer
  1. 90°
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Question 661 Mark
If the x-coordinate of a point P on the join of Q(2, 2, 1) and R(5, 1, -2) is 4, then its z-coordinate is:
  1. 2
  2. 1
  3. -1
  4. -2
Answer
  1. -1
Solution:
Suppose the point P divided the line segment joining the point Q(2, 2, 1) and R(5, 1, -2) in the ratio k : 1.
Using the section formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(5)+2}{\text{k}+2},\frac{\text{k}(1)+2}{\text{k}+1},\frac{\text{k}(-2)+1}{\text{k}+1}\Big) $
On the XY-plane, the Z-coordinate of any point is zero.
$\Rightarrow\frac{\text{k}(5)+2}{\text{k}+2}=4$
$\Rightarrow5\text{k}+2=4(\text{k}+1)$
$\Rightarrow\text{k}=2$
Now,
Z-coordinate of P $=\frac{\text{k}(-2)+1}{\text{k}+1}$ 
$\frac{2(-2)+1}{2+1}\ [\text{Substituting} \text{ k}=2]$
$=-1$
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Question 671 Mark
If a line makes 45°, 60° with positive direction of axes x and y then the angles it makes with the z-axis is:
  1. 30°
  2. 90°
  3. 45°
  4. 60°
Answer
  1. 60°
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Question 681 Mark
Choose the correct answer from the given four options. The reflection of the point $(\alpha,\beta,\gamma)$ in the xy–plane is:
  1. $(\alpha,\beta,0)$
  2. $(0,0,\gamma)$
  3. $(-\alpha,-\beta,\gamma)$
  4. $(\alpha,\beta,-\gamma)$
Answer
  1. $(\alpha,\beta,-\gamma)$
Solution:
In XY-plane, only the sign of z coordinate of the point got changed after the reflection. Therefore, the reflection of the point $(\alpha,\beta,\gamma)$ is $(\alpha,\beta,-\gamma).$
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MCQ 691 Mark
If the points $(p, 0), (0, q)$ and $(1, 1)$ are collinear, then $\frac{1}{\text{p}}+\frac{1}{\text{q}}$​ is equal to$:$
  • A
    $−1$
  • ✓
    $1$
  • C
    $2$
  • D
    $0$
Answer
Correct option: B.
$1$
As the points are collinear, the slope of the line joining
any two points, should be same as the slope of the line joining two other points.
Slope of the line passing through points $(x_1​, y_1​)$ and $\text{x}_2,\text{y}_2=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}$
So, slope of the line joining $(p, 0), (0, q) =$ Slope of the line joining
$(0, q)$ and $(1, 1)$
$\frac{\text{q}-0}{0-\text{p}}=\frac{1-\text{q}}{1-\text{p}}$
$-\frac{\text{q}}{\text{p}}=1-\text{q}$
Dividing both sides by $q,$
$-\frac{1}{\text{p}}=\frac{1}{\text{q}}-1$
$\Rightarrow\frac{1}{\text{p}}+\frac{1}{\text{q}}=1$
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Question 701 Mark
The lines $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ and $\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$ are:
  1. Coinicident.
  2. Skew.
  3. Intersecting.
  4. Parallel.
Answer
  1. Coincident.
Solution:
The equation of the given lines are
$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}\dots(1)$
$\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$
$=\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}\dots(2)$
Thus, the two lines are parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and pass through the points (0, 0, 0) and (1, 2, 3).
Now,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\vec{0}$ $\big[\because\vec{\text{a}}\times\vec{\text{a}}=\vec{0}\big]$
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Question 711 Mark
If P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) are collinear, then R divided PQ in the ratio:
  1. 3 : 2 internally
  2. 3 : 2 externally
  3. 2 : 1 internally
  4. 2 : 1 externally
Answer
  1. 3 : 2 externally
Solution:
Suppose the point R divides PQ in the ratio $\lambda:1$.
Coordinates of R are $\Big(\frac{5\lambda+3}{\lambda+1},\frac{4\lambda+2}{\lambda+1},\frac{-6\lambda-4}{\lambda+1}\Big)$.
But the coordinates of R are (9, 8, -10).
$\therefore\frac{5\lambda+3}{\lambda+1}=9,\frac{4\lambda+2}{\lambda+1}=8$ and $\frac{-6\lambda-4}{\lambda+1}=-10$
From each of these equations, we get
$\lambda=-\frac{3}{2}$
$\therefore$ R divided PQ in the ratio 3 : 2 externally.
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Question 721 Mark
If a line makes angles $\alpha,\beta,\gamma$ with the axis then $\cos 2\alpha+ \cos 2\beta +\cos 2\gamma=$
  1. -2
  2. -1
  3. 1
  4. 2
Answer
  1. -1
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Question 731 Mark
Choose the correct answer from the given four options.
The distance of the plane $\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=1$ from the origin is:
  1. 1.
  2. 7.
  3. $\frac{1}{7}.$
  4. None of these.
Answer
  1. 1.
Solution:
The general equation of a plane in vector form is given by $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$
Where d is the distance of the plane from the origin.
Comparing $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$ and $\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=1,$ we get
Therefore, the distance of the plane $\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=1$ from the origin is 1.
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Question 741 Mark
What are the direction ratios of normal to the plane 2x - y + 2z + 1 = 0:
  1. (2, -1, 2)
  2. $\big(1,\frac{1}{2},1\big)$
  3. (1, -2, 1)
  4. None of the above
Answer
  1. (2, -1, 2)
Solution:
Direction ratios of normal to the plane ax + by + cz + d = 0, are
a, b, c. So, here in the question the given plane is 2x - y + 2z + 1
= Thus, the direction ratios are 2, -1, 2
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Question 751 Mark
The direction cosines l, m, n of two lines are connected by the relations l + m + n = 0, lm = 0, then the angle between them is:
  1. $\frac{\pi}{3}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{2}$
  4. $0$
Answer
  1. $\frac{\pi}{3}$
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MCQ 761 Mark
Find the value of $p$ for which the points $(−5, 1), (1, p)$ and $(4, −2)$ are collinear.
  • A
    $1$
  • B
    $0$
  • ✓
    $−1$
  • D
    $2$
Answer
Correct option: C.
$−1$
The given points are $A(−5, 1), B(1, p)$ and $C(4, −2)$
We have $(x_{1​ }= −5, y_{1 ​}​= 1),(x_2 ​= 1, y_2​ = p)$ and $(x_3​ = 4, y_{3​} = −2)$
The given points $A, B$ and $C$ are collinear
Therefore, $x_1​(y_2​ − y_3​) + x_2​(y_{3​ }− y_{1​}) + x_3​(y_1 ​− y_2​) = 0$
$\Rightarrow (−5)⋅(p + 2) + 1⋅(−2−1) + 4⋅(1 − p) = 0$
$\Rightarrow (−5p − 10 − 3 + 4 − 4p) = 0$
$\Rightarrow −9p = −9$
$\Rightarrow p = −1$
Hence$, p = −1$
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Question 771 Mark
If a line has the direction ratio 18, 12, 4, then its direction cosines are:
  1. $\frac{9}{11},\frac{6}{11},\frac{2}{11}$
  2. $\frac{9}{13},\frac{6}{13},\frac{2}{13}$
  3. $\frac{9}{7},\frac{6}{7},\frac{2}{7}$
  4. None of these
Answer
  1. $\frac{9}{11},\frac{6}{11},\frac{2}{11}$
Solution:
Dr's of the line are : 18, 12, 4
$\text{Dc's}=\frac{18}{\sqrt{18^2+12^2+4^2}},\frac{12}{\sqrt{18^2+12^2+4^2}},\frac{4}{\sqrt{18^2+12^2+4^2}}$
$=\frac{18}{22},\frac{12}{22},\frac{4}{22}$
$=\frac{9}{11},\frac{6}{11},\frac{2}{11}$
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Question 781 Mark
The direction cosines of any normal to the xy plane are:
  1. 1, 0 ,0
  2. 0, 1, 0
  3. 1, 1, 0
  4. 1, 1, 0
Answer
  1. 1, 1, 0
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Question 791 Mark
A plane passing through (−1, 2, 3) and whose normal makes equal angle with the coordinate axes is:
  1. x + y + z + 4 = 0
  2. x − y + z + 4 = 0
  3. x + y + z − 4 = 0
  4. x + y + z = 0
Answer
  1. x + y + z − 4 = 0
Solution:
Since normal makes equal angles with coordinate axis.
So, it intercept with all the axis will be same. So equation of plane will be 
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}+\frac{\text{x}}{\text{a}}=1$
⇒ x + y + z = a
Now, it passes through (−1, 2, 3), so
−1 + 2 + 3 = a
⇒ a = 4
⇒ x + y + z − 4 = 0
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Question 801 Mark
If the three points A(1, 6), B(3, −4) and C(x, y) are collinear, then the equation satisfying by x and y is:
  1. 5x + y − 11 = 0
  2. 5x + 13y + 5= 0
  3. 5x − 13y + 5 = 0
  4. 13x − 5y + 5 = 0
Answer
  1. 5x + y − 11 = 0
Solution:
Since, the points A(1, 6), B(3, −4) and C(x, y) are colinear
$\therefore\begin{vmatrix}1&6&1\\3&-4&1\\\text{x}&\text{y}&1\end{vmatrix}=0$
⇒ 1(−4−y) −6(3 − x) + 1(3y + 4x) = 0
⇒ 10x + 2y − 22 = 0
⇒ 5x + y − 11 = 0
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Question 811 Mark
Choose the correct answer from the given four options.
Distance of the point $(\alpha,\beta,\gamma)$ from y-axis is:
  1. $\beta$
  2. $|\beta|$
  3. $|\beta|+|\gamma|$
  4. $\sqrt{\text{a}^2+\gamma^2}$
Answer
  1. $\sqrt{\text{a}^2+\gamma^2}$
Solution:
Required distance $=\sqrt{(\alpha-0)^2+(\beta-\beta)^2+(\gamma-0)^2}$
$=\sqrt{\alpha+\gamma^2}$
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Question 821 Mark
The direction cosines of the line passing through P(2, 3, -1) and the origin are:
  1. $\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
  2. $\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
  3. $\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
  4. $\frac{2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{-1}{\sqrt{14}}$
Answer
  1. $\frac{-2}{\sqrt{14}},\frac{-3}{\sqrt{14}},\frac{1}{\sqrt{14}}$
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Question 831 Mark
A plane meets the coordinate axes at A, B, and C such that the centroid of $\triangle{\text{ABC}}$ is the point (a, b, c) if
the eqution of the plane is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=\text{k}$,then k =
  1. 1
  2. 2
  3. 3
  4. None of these
Answer
  1. 3
Solution:
Let and be the interceots of the given plane on the coordinate axes.
Then, the plane meets the coordinate axes at
$\text{A}(\alpha,0,0),\text{B}(0,\beta,0)$ and $\text{C}=(0,0,\gamma)$
Given that the centroid of the triangle = (a, b, c)
$\Rightarrow\Big(\frac{\alpha+0+0}{3},\frac{0+\beta+0}{3},\frac{0+0+\gamma}{3}\Big)=(\text{a}+\text{b}+\text{c})$
$\Rightarrow\Big(\frac{\alpha}{3},\frac{\beta}{3},\frac{\gamma}{3}\Big)=(\text{a},\text{b},\text{c})$
$\Rightarrow\frac{\alpha}{3}=\text{a},\frac{\beta}{3}=\text{b},\frac{\gamma}{3}=\text{c}$
$\Rightarrow\alpha=3\text{a},\beta=3\text{b},\gamma=3\text{c}\ .....(1)$
Equation of the plane whose intercepts on the coordinate axes are $\alpha,\beta$ and $\gamma$ is
$\frac{\text{x}}{\alpha}+\frac{\text{y}}{\beta}+\frac{\text{z}}{\gamma}=1$
$\Rightarrow\frac{\text{x}}{3\alpha}+\frac{\text{y}}{3\beta}+\frac{\text{z}}{3\gamma}=1\ [\text{From (1)}]$
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=3$
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Question 841 Mark
The direction ratios of the line 6x - 2 = 3y + 1 = 2z - 2 are:
  1. $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
  2. $\frac{1}{\sqrt{14}},\frac{12}{\sqrt{14}},\frac{3}{\sqrt{14}}$
  3. $1, 2, 3$
  4. None of these
Answer
  1. $1, 2, 3$
Solution:
6x - 2 = 3y + 1 = 2x - 2
$6\big(\text{x}-\frac{2}{6}\big)=3(\text{y}+\frac{1}{3}\big)=2(\text{x}-\frac{2}{2}\big)$
$\frac{\big(\text{x}-\frac{1}{3}\big)}{1}=\frac{\big(\text{y}+\frac{1}{3}\big)}{2}=\frac{(\text{x}-1)}{3}$
Line will be passing through the poits, $\Big(\frac{1}{3},-\frac{1}{3},1\Big)$
and parallel to the line having direction ratios is 1, 2, 3
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Question 851 Mark
The angle made by line $\text{r}[\cos\theta−3​\sin\theta]=5 $ with initial line is:
  1. 30°
  2. 45°
  3. 60°
  4. 90°
Answer
  1. 30°
Solution:
Given equation
$\text{r}[\cos\theta−3​\sin\theta]=5 $
$\text{x}−\sqrt{3}\text{​y}=5$
Slope of the line is $\tan\theta=\frac{1}{\sqrt{3}}​$
$\Rightarrow\theta=30^\circ$
Hence, the line makes an angle of 30° with the initial line.
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Question 861 Mark
The vector equation of the line passing through the point (-1, 5, 4) and perpendicular to the plane z = 0 is:
  1. $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}+\lambda(\hat{\text{i}}+\hat{\text{j}})$
  2. $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
  3. $\vec{\text{r}}=\hat{\text{i}}-5\hat{\text{j}}-4\hat{\text{k}}+\lambda\hat{\text{k}}$
  4. $\vec{\text{r}}=\lambda\hat{\text{k}}$
Answer
  1. $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
Solution:
Given,
a = (-1, 5, 4)
b = (0, 0, 1) [$\therefore$ 1 to plone z]
We know that,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})+\lambda\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$
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Question 881 Mark
The direction ratios of the line joining the points A(4, −3, 7) and B(1, 3, 5) are:
  1. (5, 0, 12)
  2. (3, −6, 2)
  3. (5, 6, 2)
  4. $\Big(\frac{5}{2},0,6\Big)$
Answer
  1. (3, −6, 2)
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Question 891 Mark
The direction ratios of the normal to the plane 7x + 4y - 2z + 5 = 0 are:
  1. 7, 4, -2
  2. 7, 4, 5
  3. 7, 4, 2
  4. 4, -2, 5
Answer
  1. 7, 4, -2
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Question 901 Mark
If  (0, 0),(a, 0)  and  (0, b)  are collinear, then:
  1. ab = 0
  2. a = b
  3. a = −b
  4. a - b = c
Answer
  1. ab = 0
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Question 911 Mark
The plane x + y = 0:
  1. Is parallel to z-axis
  2. Is perpendicular to z-axis
  3. Passes through z-axis
  4. None of these
Answer
  1. Passes through z-axis
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Question 921 Mark
The locus of xy + yz = 0 is:
  1. A pair of st. lines
  2. A pair of parallel lines
  3. A pair of parallel planes
  4. A pair of perpendicular planes
Answer
  1. A pair of perpendicular planes
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Question 931 Mark
If $\alpha,\beta,\gamma$ are the angle which a half ray makes with the positive directions of the axis then $\sin^2\alpha + \sin^2\beta + \sin^2\gamma =$
  1. 1
  2. 2
  3. 0
  4. -1
Answer
  1. 2
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Question 941 Mark
The equation of the plane through the origin and parallel to the plane 3x - 4y + 5z + 6 = 0:
  1. 3x - 4y - 5z - 6 = 0
  2. 3x - 4y + 5z + 6 = 0
  3. 3x - 4y + 5z = 0
  4. 3x + 4y - 5z + 6 = 0
Answer
  1. 3x - 4y + 5z = 0
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Question 951 Mark
The points (k − 1, k + 2), (k, k + 1), (k + 1, k) are collinear for:
  1. any value of k
  2. k = −21​ only
  3. no value of k
  4. integral values of k only
Answer
  1. any value of k
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Question 961 Mark
The three points ABC have position vectors (1, x, 3), (3, 4, 7) and (y, -2, -5) are collinear then (x, y):
  1. (2, -3)
  2. (-2, 3)
  3. (-2, -3)
  4. (2, 3)
Answer
  1. (2, -3)
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Question 971 Mark
The equation of the plane passing through (2, −3, 1) and is normal to the line joining the points (3, 4, −1) and (2, −1, 5) is given by:
  1. x + 5y − 6z + 19 = 0
  2. x − 5y + 6z − 19 = 0
  3. x + 5y + 6z + 19 = 0
  4. x − 5y − 6z − 19 = 0
Answer
  1. x + 5y − 6z + 19 = 0
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Question 981 Mark
A line OP where O = (0, 0, 0) makes equal angles with ox, oy, oz. The point on OP, which is at a distance of 6 units from O is:
  1. $\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$
  2. $\big(2\sqrt{3},-2\sqrt{3},2\sqrt{3}\big)$
  3. $-\big(2\sqrt{3},-2\sqrt{3},2\sqrt{3}\big)$
  4. $-\big(6\sqrt{3},-6\sqrt{3},6\sqrt{3}\big)$
Answer
  1. $\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$
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Question 991 Mark
The direction ratios of two lines AB, AC are 1, -1, -1 and 2, -1, 1. The direction ratios of the normal to the plane ABC are:
  1. 2, 3, −1
  2. 2, 2, 1
  3. 3, 2, −1
  4. −1, 2, 3
Answer
  1. 2, 3, −1
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Question 1001 Mark
If the direction ratios of two lines are given by 3lm - 4ln + mn = 0 and l + 2m + 3n = 0, then the angle between the lines is:
  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{2}$
Answer
  1. $\frac{\pi}{2}$
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