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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 4 of 4

M.C.Q (1 Marks)

Question 1511 Mark
If the lines $\text{ x - }\frac{2}{1} =\text{y}-\frac{2}{1} =\text{z}-\frac{4}{\text{k}} $ and $\text{x}-\frac{1}{\text{k}} = \text{y}-\frac{4}{2} = \text{z}-\frac{5}{1} $ are coplanar, then k can have:
  1. Exactly two values
  2. Exactly three values
  3. Exactly one value
  4. Any value
Answer
  1. Exactly two values
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Question 1521 Mark
Distance of the point $(\alpha, \beta, \gamma)$ from y-axis is:
  1. $\beta$
  2. $\mid\beta\mid$
  3. $\mid\beta+\gamma\mid$
  4. $\sqrt{\alpha^2+\gamma^2}$
Answer
  1. $\sqrt{\alpha^2+\gamma^2}$
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Question 1531 Mark
What are the DR's of vector parallel to (2, −1, 1) and (3, 4, −1)?
  1. (1, 5, −2)
  2. (−2, −5, 2)
  3. (−1, 5, 2)
  4. (−1, −5, −2)
Answer
  1. (1, 5, −2)
Solution:
Required DR's are (3 − 2, 4 + 1, −1−1) ie, (1, 5, −2)
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Question 1541 Mark
The equation of the line passing through the points $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:
  1. $\vec{\text{r}}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\lambda\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
  2. $\vec{\text{r}}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
  3. $\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
  4. $\text{None of these}$
Answer
  1. $\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
Solution:
Equation of the line passing through the points having position vectors
$\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:
$\vec{r}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big\{\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)-\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)\big\},$ where t is a parameter
$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
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Question 1551 Mark
A parallelopiped is formed by planes drawn through the point (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes. The length of a diagonal of the parallelopiped is:
  1. $7$
  2. $\sqrt{38}$
  3. $\sqrt{155}$
  4. $\text{none of these}$
Answer
  1. 7
Solution:
The given point (2, 3, 5) and (5, 9, 7) are two diagonally opposite vertices of the parallelopiped as all of theire coordinates are different.
$\therefore$ Edges of the paralleloppiped
= |2 - 5|, |3 - 9| and |5 - 7|
=3, 6 and 2.
Now,
Length of the diagonal of the parallelopiped
$=\sqrt{3^2+6^2+2^2}$
$=\sqrt{9+36+4}$
$=\sqrt{49}$
$=7$
Hence, length of the diagonal of the parallelepiped formed by the planes
Parallel to coordinate planes and drawn through point (2, 3, 5)and (5, 9, 7) is 7 units.
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MCQ 1561 Mark
The acute angle between the planes $2x - y + z = 0$ and $x + y + 2z = 3$ is:
  • A
    $45^\circ$
  • ✓
    $60^\circ$
  • C
    $30^\circ$
  • D
    $75^\circ$
Answer
Correct option: B.
$60^\circ$
We know that the angle between the planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}$
So, the angle between $2x - y + z = 0$ and $x + y + 2x = 3$ is given by
So, $\cos\theta=\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}}$
$=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}$
$=\frac{3}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}$
$\Rightarrow\theta\cos^{-1}\Big(\frac{1}{2}\Big)=60^\circ$
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MCQ 1571 Mark
The eqution of the plane through the line $x + y + 3 = 0 = 2x - y + 3z + 1$ and parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$ is:
  • ✓
    $x - 5y + 3z = 7$
  • B
    $x - 5y + 3z = -7$
  • C
    $x + 5y + 3z = 7$
  • D
    $x + 5y + 3z = -7$
Answer
Correct option: A.
$x - 5y + 3z = 7$
The equation of the plane passing though the line of intersection of the given planes is
$\text{x}+\text{y}+\text{z}+3+\lambda(2\text{x}-\text{y}+3\text{z}+1)=0$
$(1+2\lambda)\text{x}+(1-\lambda)\text{y}+(1+3\lambda)\text{z}+3+\lambda=0\ ....(1)$
This plane is parallel to the line $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}.$
It means that this line is perpendicular to the normal of the plane $(1).$
$\Rightarrow1(1+2\lambda)\text{x}+2(1-\lambda)+3(1+3\lambda)=0($Because $a_1a_2 + b_1b_2 + c_1c_2 = 0)$
$\Rightarrow1+2\lambda+2-2\lambda+3+9\lambda=0$
$\Rightarrow9\lambda+6=0$
$\Rightarrow\lambda=\frac{-2}{3}$
Substituting this in $(1),$ we get
$\Big(1+2\Big(\frac{-2}{3}\Big)\Big)\text{x}+\Big(1-\Big(\frac{-2}{3}\Big)\Big)\text{y}+\Big(1+3\Big(\frac{-2}{3}\Big)\Big)\text{z}+3+\Big(\frac{-2}{3}\Big)=0$
$\Rightarrow -x + 5y - 3z + 7 = 0$
$\Rightarrow x - 5y + 3z = 7$
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Question 1581 Mark
The direction ratios of the line of intersection of the planes 3x + 2y - z = 5 and x - y + 2z = 3 are:
  1. 3, 2, -1
  2. -3, 7, 5
  3. 1, -1, 2
  4. -11, 4, -5
Answer
  1. -3, 7, 5
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MCQ 1591 Mark
The equation of the plane through the intersection of the planes $x + 2y + 3z = 4$ and $2x + y - z = -5$ and perpendicular to the plane $5x + 3y + 6z + 8 = 0$ is:
  • A
    $7x - 2y + 3z + 81 = 0$
  • B
    $23x + 14y - 9z + 48 = 0$
  • C
    $51x - 15y - 50z + 173 = 0$
  • ✓
    None of these
Answer
Correct option: D.
None of these
The eqution of the plane passing through the line of intersection of the given planes is
$\text{x}+2\text{y}+3\text{z}-4+\lambda(2\text{x}+\text{y}-\text{z}+5)=0$
$(1+2\lambda)\text{x}+(2+\lambda)\text{y}+6(3-\lambda)\text{z}-4+5\lambda=0\ ....(1)$
This plane is perpendicular to $5x + 3y + 6z + 8 = 0.$
So, $5(1+2\lambda)+3(2+\lambda)+6(3-\lambda)=0 ($Because $a_1a_2 + b_1b_2 + c_1c_2 = 0)$
$\Rightarrow5+10\lambda+6+3\lambda+18-6\lambda=0$
$\Rightarrow7\lambda+29=0$
$\Rightarrow\lambda=\frac{-29}{7}$
Substituting this in $(1),$ we get
$\Big(1+2\Big(\frac{-29}{7}\Big)\Big)\text{x}+\Big(2+\Big(\frac{-29}{7}\Big)\Big)\text{y}+\Big(3+\frac{29}{7}\Big)\text{z}-4+5\Big(\frac{-29}{7}\Big)=0$
$\Rightarrow 51x + 15y - 50z + 173 = 0.$
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Question 1601 Mark
Choose the correct answer from the given four options.
The plane 2x – 3y + 6z – 11 = 0 makes an angle $\sin^{-1}(\alpha)$ with x-axis. The value of $\alpha$ is equal to:
  1. $\frac{\sqrt{3}}{2}$
  2. $\frac{\sqrt{2}}{3}$
  3. $\frac{2}{7}$
  4. $\frac{3}{7}$
Answer
  1. $\frac{2}{7}$
Solution:
We are given that, 2x - 3y + 6z - 11 = 0 makes angle $\sin^{-1}(\alpha)$ with x-axis.
 The equation of plane 2x - 3y + 6z - 11 = 0 in vector form is given by $\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}})=11$
$\therefore\vec{\text{b}}=(\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$ and $\vec{\text{n}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$
We know that, $\sin\theta=\frac{|\vec{\text{b}}\cdot\vec{\text{n}}|}{|\vec{\text{b}|}\cdot|\vec{\text{n}}|}$
$=\frac{\big|(\vec{\text{i}})\cdot(2\vec{\text{i}}-3\vec{\text{j}}+6\vec{\text{k}})\big|}{\sqrt{1}\sqrt{4+9+36}}$
$=\frac{2}{7}$
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Question 1611 Mark
The vector equation r = i − 2j − k + t(6j − k) represents a straight line passing through the points:
  1. (0, 6, −1) and (1, −2, −1)
  2. (0, 6, −1) and (−1, −4, −2)
  3. (1, −2, −1) and (1, 4, −2)
  4. (1, −2, −1) and (0, −6, 1)
Answer
  1. (1, −2, −1) and (1, 4, −2)
Solution:
 
Cartesian representation of the given line is,
$\frac{\text{x}-1}{0}=\frac{\text{y}+2}{6}=\frac{\text{z}+1}{-1}=\text{t}$
So any point on the given line is of the form (1, 6t − 2, − t − 1) where t can be any real numbers
So for t = 0 and 1 the corresponding points are (1, −2, −1) and (1, 4, −2)
You can check other options does not satisfy above point for any t.
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Question 1621 Mark
The distance of the point (-1, -5, -10) from the point of intersection of the line $\vec{\text{r}}.=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$ and the plane $\vec{\text{r}}.=(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$ is:
  1. 9
  2. 13
  3. 17
  4. None of these
Answer
  1. 13
Solution:
Given equation of line is
$\vec{\text{r}}.=(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(3\hat{\text{i}}+4\hat{\text{j}}+12\hat{\text{k}})$
$\vec{\text{r}}.=(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+12\lambda)\hat{\text{k}}$
The coordinates of any point on this line are of the from
$(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+12\lambda)\hat{\text{k}}$ or $(2+3\lambda,-1+4\lambda,2+12\lambda)$
Scince this point lies on the plane $\vec{\text{r}}.(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5,$
$\Big[(2+3\lambda)\hat{\text{i}}+(-1+4\lambda)\hat{\text{j}}+(2+12\lambda)\hat{\text{k}}\Big].(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5,$
$\Rightarrow2+3\lambda+1-4\lambda+2+12\lambda-5=0$
$\Rightarrow\lambda=0$
So, the coordinates of the point are
$(2+3\lambda,-1+4\lambda,2+2\lambda)$
$=(2+0,-1+0,2+0)$
$=(2, -1,2)$
Distance between (2, -1, 2) and (-1, -5, -10)
$=\sqrt{(1-2)^2+(-5+1)^2+(-10-2)^2}$
$=\sqrt{9+16+144}$
$=13 \text{ units}$
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Question 1631 Mark
If a line makes angle $\alpha,\beta$ and $\gamma$ with the axes respectively, then $\cos2\alpha+\cos2\beta+\cos2\gamma=$
  1. -2
  2. -1
  3. 1
  4. 2
Answer
  1. -1
Solution:
If a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then
$\cos2\alpha+\cos2\beta+\cos2\gamma=1\dots(1)$
We have
$\cos2\alpha+\cos2\beta+\cos2\gamma$
$=2\cos^2\alpha-1+2\cos^2\beta-1+2\cos^2\gamma-1$ $\big[\because\cos2\theta=2\cos^2\theta-1\big]$
$=2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)-3$ [From (1)]
$=2(1)-3$
$=-1$
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MCQ 1641 Mark
If the direction cosine of a directed line be $a, 3a, 7a$ then $a =$
  • ✓
    $\underline{+}\frac{1}{59}$
  • B
    $\underline{+}\frac{1}{9}$
  • C
    $\underline{+}\frac{2}{7}$ 
  • D
    None of these
Answer
Correct option: A.
$\underline{+}\frac{1}{59}$
Give, $a, 3a, 7a$ be the direction cosines of a directed line.
Then from the property of direction cosines we get
$a^2 + (3a)^2+ (7a)^2 = 1$ or
$59a^2 = 1$ or
$\text{a}=\underline{+}\frac{1}{\sqrt{59}}$
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Question 1651 Mark
Choose the correct answer from the given four options.
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:
  1. 9 sq. units.
  2. 18 sq. units.
  3. 27 sq. units.
  4. 81 sq. units.
Answer
  1. 9 sq. units.
Solution:
We have, A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2)
$\therefore\overrightarrow{\text{AB}}=(2-0)\hat{\text{i}}+(3-4)\hat{\text{j}}+(-1-1)\hat{\text{k}}$
$=2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}$
$\overrightarrow{\text{BC}}=(4-2)\hat{\text{i}}+(5-3)\hat{\text{j}}+(0-0)\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\overrightarrow{\text{CD}}=(2-4)\hat{\text{i}}+(6-5)\hat{\text{j}}+(2-0)\hat{\text{k}}$
$=-2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\overrightarrow{\text{DA}}=(0-2)\hat{\text{i}}+(4-6)\hat{\text{j}}+(1-2)\hat{\text{k}}$
Thus quadrilateral formed is parallelogram.
Area of quadrilateral ABCD
$=\big|\overrightarrow{\text{AB}}\times\overrightarrow{\text{BC}}\big|=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\2&-1&-2\\2&2&1 \end{vmatrix}$
$=|3\vec{\text{i}}-6\vec{\text{j}}+6\vec{\text{k}}|$
$=\sqrt{9+36+36}$
$=9\text{ sq.units}$
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Question 1661 Mark
A line makes an angle $\alpha,\beta,\gamma$ with the X, Y, Z axes. Then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=$
  1. 1
  2. 2
  3. $\dfrac{3}{2}$
  4. 4
Answer
  1. 2
Solution:
For a vector.
$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=1$
$1-\sin^2(\alpha)+1-\sin^2(\beta)+1-\sin^2(\gamma)=1$
$\sin^2(\alpha)\sin^2(\beta)+\sin^2(\gamma)=3-1=2$
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Question 1671 Mark
The Image of the point (2, -1, 5) in the plane $\vec{\text{r}},\hat{\text{i}}=0$ is:
  1. (-2, -1, 5)
  2. (2, 1, -5)
  3. (-2, 1, -5)
  4. (2, 0, 0)
Answer
  1. (-2, -1, 5)
Solution:
Equation of plane is r.i = 0
i.e. x = 0
It is equation of Y-Z plane
Let PQ be the line Perpendicular to the plane from (2, -1, 5)
Also line is perpendicular to plane so direction ratios of line will be that of the DR's of plane equation of line will be:
$\frac{(\text{x}-2)}{(1)} = \frac{(\text{x}-b)}{0} = \frac{(\text{x}-\text{c})}{0} = \text{k say}$
General points of line PQ will be
x = k + 2
y = -1
z = 5
Also, this line intersect the plane
so, foot of perpendicular will be
(k + 2) = 0
k = -2
Hence foot of perpendicular will be (0, -1, 5)
let coordinates of image is (e, f, g)
By mid-point theorem
$0 = \frac{(2 + \text{e})}{2} \Rightarrow \text{e} = -2$
$-1 = \frac{(-1 + \text{f})}{2}\Rightarrow\text{f} = -1$
$5 = \frac{(5 + \text{g})}{2}\Rightarrow\text{g} = 5$
So, coordinates of image is
(-2, -1, 5)
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Question 1681 Mark
If the direction cosines of a line are $\Big(\frac{1}{\text{c}},\frac{1}{\text{c}},\frac{1}{\text{c}}\Big)$ then:
  1. 0 < c < 1
  2. c > 2
  3. $\text{c}=\underline{+}\sqrt{2}$
  4. None of these
Answer
  1. $\text{c}=\underline{+}\sqrt{2}$
Solution:
Since, DC′s of a line are $\Big(\frac{1}{\text{c}},\frac{1}{\text{c}},\frac{1}{\text{c}}\Big)$
$\because\text{l}^2 + \text{m}^2 + \text{n}^2 = 1$
$\because\Big(\frac{1}{\text{c}}\Big)^2+\Big(\frac{1}{\text{c}}\Big)^2+\Big(\frac{1}{\text{c}}\Big)^2=1$
$\Rightarrow1 + 1 + 1 = \text{c}^2$
$\Rightarrow\text{c}^2 = 3$
$\Rightarrow\text{c}=\underline{+}\sqrt{3}$
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Question 1691 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. Parallel.
  2. Intersecting.
  3. Skew.
  4. Coincident.
Answer
  1. coincident
Solution:
The equations of the given lines are
$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}\dots(1)$
$\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$
$\Rightarrow\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}\dots(2)$
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,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$
$=\vec{0}$ $\big[\because\vec{\text{a}}\times\vec{\text{a}}=\vec{0}\big]$
Since, the distence between the two parallel lines is 0, the given two lines are coincident lines.
Disclaimar: The answer given in the book is incorrect. This solution is created according to the question given in the book.
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Question 1701 Mark
The eqution of the plane which cute equal intercepts of unit length on the coordinate axes is:
  1. x + y + z = 1
  2. x + y + z = 0
  3. x + y - z = 1
  4. x + y + z = 2
Answer
  1. x + y + z = 1
Solution:
We know that the equation of aplane whose intercepts are a, b, c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(1)$
It is given that a = b = c
So, from (1),
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
$\Rightarrow\text{x}+\text{y}+\text{z}=\text{a}\ ....(2)$
Since it is given that the intercepts of the required plane are of unit length,
a = b = c = 1
Substituting a = 1 in (2), we get
x + y + z = 1
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Question 1711 Mark
A straight line passes through (1, -2, 3) and perpendicular to the plane 2x + 3y - z = 7. Find the direction ratios of normal to plane:
  1. < 2, 3, -1 >
  2. < 2, 3, 1 >
  3. < -1, 2, 3 >
  4. None of the above
Answer
  1. < 2, 3, -1 >
Solution:
concept: for any plane ax + by + cz + d =
0, normal vector to this plane is $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
the normal vector of the plane 2x + 3y - z = 7 is $\text{2}\hat{\text{i}}+\text{3}\hat{\text{j}}+\hat{\text{k}}$
so the direction ratios of normal to plane are < 2, 3, -1 >
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Question 1721 Mark
Find the equation of the plane passing through the points P(1, 1, 1), Q(3, -1, 2), R(-3, 5, -4):
  1. x + 2y = 0
  2. x - y - 2 = 0
  3. -x + 2y - 2 = 0
  4. x + y - 2 = 0
Answer
  1. x + y - 2 = 0
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Question 1741 Mark
The distance of the point P(a, b, c) from the x-axis is:
  1. $\sqrt{\text{b}^2+\text{c}^2}$
  2. $\sqrt{\text{a}^2+\text{c}^2}$
  3. $\sqrt{\text{a}^2+\text{b}^2}$
  4. $\text{none of these}$
Answer
  1. $\sqrt{\text{b}^2+\text{c}^2}$
Solution:
The projection of the point P(a, b, c) on the x-axis is a, (0, 0) as both Y and Z coordinates on any point on the x-axis are equal to zero.
$\therefore$ Distance of P(a, b, c) from x-axis = Distance of P(a, b, c) from a, (0, 0)
$=\sqrt{(\text{a}-\text{a})^2+(\text{b}-0)^2+(\text{c}-0)^2}$
$=\sqrt{\text{b}^2+\text{c}^2}$
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Question 1751 Mark
What are the DRs of vector parallel to (2, -1, 1) and (3, 4, -1):
  1. (1, 5, -2)
  2. (-2, -5, 2)
  3. (-1, 5, 2)
  4. (-1, -5, -2)
Answer
  1. (1, 5, -2)
Solution:
Required DRs are (3 - 2, 4 + 1, -1 - 1) ie, (1, 5, -2)
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Question 1761 Mark
If a plane passes through the point (1, 1, 1) and is perpendicular to the line $\frac{\text{x}-1}{3}=\frac{\text{y}-1}{0}=\frac{\text{z}-1}{4}$ then its perpendicular distance from the origin is:
  1. $\frac{3}{4}$
  2. $\frac{4}{3}$
  3. $\frac{7}{5}$
  4. $1$
Answer
  1. $\frac{7}{5}$
Solution:
Since the plane is perpendicular to the given line, its direction ratios are proportinal to 3, 0, 4.
So the required equation of the plane is of the form
3x + 0y + 4z + d = 0 .....(1), where d is a constant.
Since this plane passes through (1, 1, 1),
3 + 0 + 4 + d = 0
d = -7
Substituting this in (1), we get
3x + 0y + 4z -7 = 0 ......(2)
perpendicular distance of (2) from the origin
$=\frac{|3(0)+0+4(0)-7|}{\sqrt{3^2+0^2+4^2}}$
$=\frac{|0+0-7|}{\sqrt{25}}$
$=\frac{7}{5}\text{ units}$
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Question 1771 Mark
If l, m, n are the d.cs of the line joining (5, -3, 8) and (6, -1, 6) then l + m + n =
  1. $1$
  2. $\frac{1}{3}$
  3. $-1$
  4. $\frac{5}{3}$
Answer
  1. $\frac{1}{3}$
Solution:
The line joining (5, -3, 8) and (6, -1, 6) is given by the vector -i + 2j - 2k.
the direction cosines are given by. l =
$\frac{1}{\sqrt{1^2+2^2+(-2)^2}}=\frac{1}{3}, \text{m}=\frac{2}{\sqrt{1^2+2^2+(-2)^2}}=\frac{2}{3}$
$\text{n}=\frac{-2}{1^2+2^2+(-2)^2}=\frac{-2}{3}$
$\Rightarrow\text{l + m + n}=\frac{1}{3}$
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Question 1781 Mark
A line makes angles $\alpha,\beta,\gamma$ with the positive direction of the axes of reference. The value of $\cos2\alpha+\cos2\beta+\cos2\gamma$ is:
  1. 1
  2. 3
  3. -1
  4. 0
Answer
  1. -1
Solution:
$\cos^2\alpha+\cos^2\beta+\cos^2\text{r}=1\cos2\alpha+\cos^2\beta+\cos2\text{r}$
$=2\cos^2\alpha-1+2\cos^2\beta-1+2\cos^2\text{r}-1$
$=2(\cos^2\alpha+\cos^2\beta\cos{\text{r}})-3=2(1)-3=-1$
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Question 1791 Mark
Area of $\triangle\text{ABC}$ is:
  1. 45 squares units
  2. 55 squares units
  3. 65 squares units
  4. none of these
Answer
  1. none of these
Solution:
Line PA: $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}-6}{1}$
Line PB: $\frac{\text{x}-1}{-1}=\frac{\text{y}-2}{2}=\frac{\text{z}-6}{1}$
Line PC: $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{-1}=\frac{\text{z}-6}{-2}$
Then $\text{A}\Big(\frac{7}{2},-\frac{1}{2},\frac{17}{2}\Big)$
$\text{B}\Big(\frac{17}{2},-13,-\frac{3}{2}\Big)$
$\text{C}\Big(-14,\frac{19}{2},21\Big)$
Hence area of $\triangle\text{ABC}=\frac{225\sqrt{14}}{8},$ volume of tetrahedron
$\text{PABC}=\frac{125}{8}\text{cubic units}$
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Question 1801 Mark
If $\cos\alpha,\cos\beta,\cos\gamma$ are the direction cosines of a vector $\vec{\text{a}}$ then $\cos2\alpha+\cos2\beta+\cos2\gamma$ is equal to:
  1. 2
  2. 3
  3. -1
  4. 0
Answer
  1. -1
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