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MATHS M.C.Q (1 Marks)

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

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M.C.Q (1 Marks)

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Question 11 Mark
The direction cosines of the normal to the plane 2x - 3y - 6z - 3 = 0 are:
  1. $\frac{2}{7},\frac{-3}{7},\frac{-6}{7}$
  2. $\frac{2}{7},\frac{3}{7},\frac{6}{7}$
  3. $\frac{-2}{7},\frac{-3}{7},\frac{-6}{7}$
  4. None of these
Answer
  1. $\frac{2}{7},\frac{-3}{7},\frac{-6}{7}$
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Question 21 Mark
The points A(1, 1, 0), B(0, 1, 1), C(1, 0, 1) and $\text{D}\big(\frac{2}{3},\frac{2}{3},\frac{2}{3}\big)$
  1. Coplanar
  2. Non-coplanar
  3. Vertices of a parallelogram
  4. None of these
Answer
  1. Coplanar
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Question 31 Mark
The sum of the squares of sine of the angles made by the line AB with OX, OY, OZ where O is the origin is:
  1. 1
  2. 2
  3. -1
  4. 3
Answer
  1. 2
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Question 41 Mark
A point P lies on the line segment joining the points (-1, 3, 2) and (5, 0, 6). If x-coordinate of P is 2, then its z-coordinate is:
  1. $-1$
  2. $4$
  3. $\frac{3}{2}$
  4. $8$
Answer
  1. $4$
Solution:
Equation of time passing through (-1, 3, 2) and (5, 0, 6)
$\frac{\text{x}+1}{5-(-1)}=\frac{\text{y}-3}{0-3}=\frac{\text{z}-2}{6-2}$
$\frac{\text{x}+1}{6}=\frac{\text{y}-3}{-3}$
$=\frac{\text{z}-2}{4}=\text{k}$
Any point on it,
$\text{P}(6\text{k}-1,\text{3}\text{k}+3,4\text{k}+2)$
x Coordinate $=2$
$=6\text{k}-1$
$\Rightarrow\text{k}=\text{y}_2$
z Coordinate $=4\text{k}+2$
$=4\Big(\frac{1}{2}\Big)+2$
$=2+2=4$
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Question 51 Mark
The vector equation of the plane containing the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$ and the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ is:
  1. $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
  2. $\vec{\text{r}}.(\hat{\text{i}}-3\hat{\text{k}})=10$
  3. $\vec{\text{r}}.(3\hat{\text{i}}+\hat{\text{k}})=10$
  4. $\text{None of these}$
Answer
  1. $\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
Solution:
Let the direction ratio of the required plane be proportinal to a, b, c.
Scince the required plane contains the line $\vec{\text{r}}=(-2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}})+\lambda(3\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}})$
It must pass through the point (-2, -3, 4) and it should be parallel to the line.
So, the equation of the plane is
a(x + 2) + b(y + 3) + c(z - 4) = 0 ....(1) and
3a - 2b - c = 0 ....(2)
It is given that plane (1) passes through the point $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ or (1, 2, 3).
a(1 + 2) + b(2 + 3) + c(3 - 4) = 0
3a + 5b - c = 0 .......(3)
So,
Solving (1) (2) and (3), we get
$\begin{vmatrix}\text{x}+2&\text{y}+3&\text{z}-4\\3&-2&-1\\3&5&-1\end{vmatrix}=0$
$\Rightarrow7(\text{x}+2)+0(\text{y}+3)+21(\text{y}-4)=0$
$\Rightarrow\text{x}+2+3\text{z}-12=0$
$\Rightarrow\text{x}+3\text{z}=10$ or $\vec{\text{r}}.\big(\hat{\text{i}}+3\hat{\text{k}}\big)=10$
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MCQ 61 Mark
$l = m = n = 1$ represents the direction cosines of:
  • A
    $x−$axis
  • B
    $y−$axis
  • C
    $z−$axis
  • ✓
    none of these
Answer
Correct option: D.
none of these
Suppose, $\text{l, m, n}$ are direction cosines
$\Rightarrow 1^2 + m^2 + n^2 = 1$
But $ 1 = m = n = 1$
$\Rightarrow 3m^2 = 1$
$\Rightarrow 1 = m = n = \frac{1}{\sqrt3}$​
which are not direction cosines of either of the three co$-$ordinate axes.
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Question 71 Mark
Can $\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$​ be the direction cosines of any directed line?
  1. Yes
  2. No
  3. Cannot say
  4. None of these
Answer
  1. No
Solution:
No, they can not be the direction cosines of any directed line.
As the sum of square of them is not 1.
$\text{As}\Big(\frac{1}{\sqrt{3}}\Big)^2+\Big(\frac{2}{\sqrt{3}}\Big)^2+\Big(\frac{-2}{\sqrt{3}}\Big)^2$
$=\frac{1+4+4}{3}$
$=3$
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Question 81 Mark
The angle between the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$ and $\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$ is:
  1. $\cos^{-1}\big(\frac{1}{65}\big)$
  2. $\frac{\pi}{6}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{4}$
Answer
  1. $\frac{\pi}{3}$
Solution:
We have
$\frac{\text{x}-1}{1}=\frac{\text{y}-1}{1}=\frac{\text{z}-1}{2}$
$\frac{\text{x}-1}{-\sqrt{3}-1}=\frac{\text{y}-1}{\sqrt{3}-1}=\frac{\text{z}-1}{4}$
The direction ratios of the given lines are proportional to 1, 1, 2 and $-\sqrt{3}-1,\sqrt{3}-1, 4$
The given lines are parallel to vectors $\vec{\text{b}}_1=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}_2=\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big\{\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}}{\sqrt{1^2+1^2+1^2}\sqrt{\big(-\sqrt{3}-1\big)^2+\big(\sqrt{3}-1\big)^2+4^2}}$
$=\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{3}\sqrt{24}}$
$=\frac{6}{6\sqrt{2}}$
$\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\frac{\pi}{3}$
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MCQ 91 Mark
length of the $1^{er}$ from the point $(0, -1, 3)$ to the plane $2x + y - 2z + 1 = 0$ is$:$
  • A
    $0$
  • B
    $2\sqrt{3}$
  • C
    $\frac{2}{3}$
  • ✓
    $2$
Answer
Correct option: D.
$2$
$2$
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Question 101 Mark
If the diraction ratios of a line are proportional to 1, -3, 2, then its diraction cosines are:
  1. $\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
  2. $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
  3. $-\frac{1}{\sqrt{14}},\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
  4. $-\frac{1}{\sqrt{14}},-\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}$
Answer
  1. $\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
Solution:
The diraction ratios of the line are proportional to 1, -3, 2.
$\therefore$ The direction cosines of the line are
$\frac{1}{\sqrt{1^2+(-3)^2+2^2}},\frac{-3}{\sqrt{1^2+(-3)^2+2^2}},\frac{2}{\sqrt{1^2+(-3)^2+2^2}}$
$=\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$
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MCQ 111 Mark
If a line makes angles $Q_1, Q_{21}$ and $Q_3$ respectively with the coordinate axis then the value of $\cos^2 \text{Q}_{1} + \cos^2 \text{Q}_{2} + \cos^2 \text{Q}_{3}:$
  • A
    $2$
  • ✓
    $1$
  • C
    $4$
  • D
    $\frac{3}{2}$
Answer
Correct option: B.
$1$
$1$
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Question 121 Mark
The equation x² - x - 2 = 0 in three-dimensional space is represented by:
  1. A pair of parallel planes
  2. A pair of straight lines
  3. A pair of the perpendicular plane
  4. None of these
Answer
  1. A pair of parallel planes
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Question 131 Mark
If 2x + 5y - 6z + 3 = 0 be the equation of the plane, then the equation of any plane parallel to the given plane is:
  1. 3x + 5y – 6z + 3 = 0
  2. 2x - 5y - 6z + 3 = 0
  3. 2x + 5y - 6z + k = 0
  4. None of these
Answer
  1. 2x + 5y - 6z + k = 0
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Question 141 Mark
The equation of the plane passing through the points (3, 2, −1), (3, 4, 2) and (7, 0, 6) is 5x + 3y −2z = λ where λ is:
  1. 23
  2. 21
  3. 19
  4. 27
Answer
  1. 23
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Question 151 Mark
If the d.rs of two lines are 1, -2, 3 and 2, 0, 1, then the d.rs of the line perpendicular to both the given lines is:
  1. -2, 5, 4
  2. 2, -5
  3. 2, 5, -4
  4. 1, 5, -4
Answer
  1. -2, 5, 4
Solution:
OA and OB are given by (1, -2, 3), (2, 0, 1)
A line that will be perpendicular to both OA and OB can be obtained by doing the cross product of OA with OB.
Then, n = OA × OB
n = -2i + 5j + 4k
(-2, 5, 4).
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Question 161 Mark
Can $\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$ be the direction cosines of any directed line:
  1. Yes
  2. No
  3. Cannot say
  4. None of these
Answer
  1. No
Solution:
No, they can not be the direction cosines of any directed line. As the sum of square of them is not 1.As
$=\Big(\frac{1}{\sqrt{3}}\Big)^2+\Big(\frac{2}{\sqrt{3}}\Big)^2+\Big(\frac{-2}{\sqrt{3}}\Big)^2=\frac{1+4+4}{3}=3$
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Question 171 Mark
A vector parallel to the line of intersection of the plance $\vec{\text{r}}.(3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=1$ and $\vec{\text{r}}.(\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}})=2$ is:
  1. $-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
  2. $2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$
  3. $-2\hat{\text{i}}-7\hat{\text{j}}+13\hat{\text{k}}$
  4. $2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
Answer
  1. $2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$
Solution:
Let the required vector be a $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\ ....(1)$
Since the vector is parallel to the line of intersection of the given planes,
3a - b + c = 0 .....(2)
a + 4b - 2c = 0 ....(3)
Solving (2) and (3), we get
$\frac{\text{a}}{-2}=\frac{\text{b}}{7}=\frac{\text{c}}{13}$
Substituting these values in (1), we get
$-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$ which is the required vector.
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Question 181 Mark
The angle between the planes 2x - y + z = 6 and x + y + 2z = 7 is:
  1. $\frac{\pi}{4}$
  2. $\frac{\pi}{6}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{2}$
Answer
  1. $\frac{\pi}{3}$
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Question 191 Mark
The distance of the plane 2x - 3y + 6z + 7 = 0 from the point (2, -3, -1) is:
  1. 4
  2. 3
  3. 2
  4. $\frac{1}{5}$
Answer
  1. 2
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Question 201 Mark
The equation of the plane through point (1, 2, -3) which is parallel to the plane 3x - 5y + 2z = 11 is given by:
  1. 3x - 5y + 2z - 13 = 0
  2. 5x - 3y + 2z + 13 = 0
  3. 3x - 2y + 5z + 13 = 0
  4. 3x - 5y + 2z + 13 = 0
Answer
  1. 3x - 5y + 2z + 13 = 0
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Question 211 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 221 Mark
The cosines of the angle between any two diagonals of a cube is:
  1. $\frac{1}{3}$
  2. $\frac{1}{2}$
  3. $\frac{2}{3}$
  4. $\frac{1}{\sqrt{3}}$
Answer
  1. $\frac{1}{3}$
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Question 231 Mark
The xy-plane divided the line joining the point (-1, 3, 4) and (2, -5, 6)
  1. Internally in the ratio 2 : 3
  2. Externally in the ratio 2 : 3
  3. Internally in the ratio 3 : 2
  4. Externally in the ratio 3 : 2
Answer
  1. Externally in the ratio 2 : 3
Solution:
Let the XY-plane divide the line segment joining points
P(-1, 3, 4) and Q(2, -5, 6) in the ratio k : 1.
Using the section formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(2)-1}{\text{k}+1},\frac{\text{k}(-5)+3}{\text{k}+1},\frac{\text{k}(6)+4}{\text{k}+1}\Big) $
On the XY-plane, the Z-coordinate of any point is zero.
$\Rightarrow\frac{\text{k}(6)+4}{\text{k}+1}=0$
$\Rightarrow6\text{k}+4=0$
$\Rightarrow\text{k}=\frac{-2}{3}$
Thus, the XY-plane divides the line segment joining the given points in the ratio 2 : 3 externally.
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Question 241 Mark
Choose the correct answer from the given four options.
The locus represented by xy + yz = 0 is:
  1. A pair of perpendicular 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.
Solution:
We have, xy + yz = 0
⇒ xy = -yz
So, a pair of perpendicular planes.
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MCQ 251 Mark
If $l, m, n$ are the direction cosines of a line, then$:$
  • A
    $l^{2 }+ m^{2 }+ 2n^2 = 1$
  • B
    $l^{2 }+ 2m^{2 }+ n^2 = 1$
  • C
    $2l^{2 }+ m^{2 }+ n^2 = 1$
  • ✓
    $l^{2 }+ m^{2 }+ n^2 = 1$
Answer
Correct option: D.
$l^{2 }+ m^{2 }+ n^2 = 1$
$l^{2 }+ m^{2 }+ n^2 = 1$
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Question 261 Mark
For every point P(x, y, z) on the x-axis (except the origin),
  1. x = 0, y = 0, z ≠ 0
  2. y = 0, z = 0, y ≠ 0
  3. y = 0, z = 0, x ≠ 0
  4. x = y = z = 0
Answer
  1. y = 0, z = 0, x ≠ 0
Solution:
Both Y and Z coordinates on each point of the x-axis are equal to zero.
The X-coordinate on the origin is also equal to zero.
Therefore, the Y and Z coordinates on each point of the x-axis, except the origin, are equal to zero,
While the X-coordinate is non-zero.
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Question 271 Mark
The equation of the plane parallel to the lines x - 1 = 2y - 5 = 2z and 3x = 4y - 11 = 3z -4 and passing through the point (2, 3, 3) is:
  1. x - 4y + 2z + 4 = 0
  2. x + 4y + 2z + 4 = 0
  3. x - 4y + 2z - 4 = 0
  4. None of these
Answer
  1. x - 4y + 2z + 4 = 0
Solution:
Let a, b, c be the dirction ratios of the required plane.
The given line equation can be rewritten as
$\frac{\text{x}-1}{1}=\frac{\text{y}-\frac{5}{2}}{\frac{1}{2}}=\frac{\text{z}-0}{\frac{1}{2}}\ .....(1)$
$\frac{\text{x}-0}{\frac{1}{3}}=\frac{\text{y}-\frac{11}{4}}{\frac{1}{4}}=\frac{\text{z}-\frac{4}{3}}{\frac{1}{3}}\ .....(2)$
Since the required plane is parallel to the lines (1) and (2),
$\text{a}+\frac{\text{b}}{2}+\frac{\text{c}}{2}=0\Rightarrow2\text{a}+\text{b}+\text{c}=0....(3)$
$\frac{\text{a}}{1}=\frac{\text{b}}{-4}=\frac{\text{c}}{2}=0\Rightarrow4\text{a}+3\text{b}+4\text{c}=0....(4)$
Solving (3) and (4) using cross-multiplication method, we get
$\frac{\text{a}}{1}=\frac{\text{b}}{-4}=\frac{\text{c}}{2}=\lambda\text{(say)}$
$\Rightarrow\text{a}=\lambda,\text{b}=-4\lambda,\text{c}=2\lambda$
Now, the eqution of the plane whose direction ratios are $\lambda,-4\lambda,2\lambda$ and passing through the point.
$\lambda(\text{x}-2)+(-4\lambda)(\text{y}-3)+2\lambda(\text{z}-3)=0$
$\Rightarrow\text{x}-4\text{y}+2\text{z}+4=0$
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Question 281 Mark
Direction cosines of ray from P(1, −2, 4) to Q(−1, 1, −2) are:
  1. −2, 3, −6
  2. 2, −3, 6
  3. 2, 3, 6
  4. $\frac{-2}{7},\frac{3}{7},\frac{-6}{7}$
Answer
  1. $\frac{-2}{7},\frac{3}{7},\frac{-6}{7}$
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
$\bigg(\frac{2}{\sqrt{2^2+3^2+6^2}},\frac{3}{\sqrt{2^2+3^2+6^2}},\frac{-6}{\sqrt{2^2+3^2+6^2}}\bigg)=\Big(\frac{-2}{7},\frac{3}{7},\frac{-6}{7}\Big).$
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MCQ 291 Mark
Choose the correct answer from the given four options. If the directions cosines of a line are $k, k, k,$ then$:$
  • A
    $\text{k}>0$
  • B
    $0<\text{k}<1$
  • C
    $\text{k}=1$
  • ✓
    $\text{k}=\frac{1}{\sqrt{3}}\text{ or }-\frac{1}{\sqrt{3}}$
Answer
Correct option: D.
$\text{k}=\frac{1}{\sqrt{3}}\text{ or }-\frac{1}{\sqrt{3}}$
Since, direction cosines of a line are $k, k,$ and $k.$
$\therefore l = k, m = k$ and $n = k$
We know that, $l^2 + m^2 + n^2 = 1$
$\Rightarrow k^2 + k^2 + k^2 = 1$
$\text{k}^2=\frac{1}{3}$
$\therefore\text{k}=\pm\frac{1}{\sqrt{3}}$
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Question 301 Mark
The straight line $\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$ is:
  1. Parallel to x-axis
  2. Parallel to y-axis
  3. Parallel to z-axis
  4. Perpendicular to z-axis
Answer
  1. Perpendicular to z-axis
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Question 311 Mark
If O is the origin, OP = 3 with direction ratios proportional to -1, 2, -2 then the coordinates of P are:
  1. $(-1, 2,-2)$
  2. $(1, 2, 2)$
  3. $\Big(\frac{-1}{9},\frac{2}{9},\frac{-2}{9}\Big)$
  4. $(3,6,-9)$
Answer
  1. $(-1, 2,-2)$
Solution:
Let the coordinates of P be (x, y, z). Then,
Direction ratios of OP = Coordinates of P-Coordinates of O-1, 2, 2 = (x - 0), (y - 0), (z - 0)
Thus, coordinates of P are (-1, 2, -2).
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Question 321 Mark
Direction ratio of line joining (2, 3, 4) and (-1, -2, 1), are:
  1. (-3, -5, -3)
  2. (-3, 1, -3)
  3. (-1, -5, -3)
  4. (-3, -5, 5)
Answer
  1. (-3, -5, -3)
Solution:
The direction ratio of the line joining A(2, 3, 4) and B(-1, -2, 1), are.
= (-1 - 2), (-2 - 3), (1 - 4)
= (-3, -5, -3)
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Question 331 Mark
The distance of the points (2, 1, -1) from the plane x - 2y + 4z - 9 is:
  1. $\frac{\sqrt{31}}{21}$
  2. $\frac{13}{21}$
  3. $\frac{13}{\sqrt{21}}$
  4. $\sqrt{\frac{\pi}{2}}$
Answer
  1. $\frac{13}{\sqrt{21}}$
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MCQ 341 Mark
A line makes the same angle $\theta$ with each of thex and $z$ axis. If the angle $\beta$ which it makes with $y-$axis is such that $\sin^2\beta=3\sin^2\theta$ then $\cos^2\theta$ equals:
  • ✓
    $\frac{3}{5}$
  • B
    $\frac{1}{5}$
  • C
    $\frac{2}{3}$
  • D
    $\frac{2}{5}$
Answer
Correct option: A.
$\frac{3}{5}$
If a line makes the angle $\alpha,\beta,\gamma$ with $x, y, z$ axix respectively then
$l^2 + m^2 + n^2 = 1$
$\Rightarrow 2l^2 + m^2 = 1$ or $2n^2 + m^2 = 1$
$\Rightarrow2\cos^2\theta=1-\cos^2\beta (\alpha=\gamma=\theta)$
$2\cos^2\theta=\sin^2\beta$
$\Rightarrow2\cos^2\theta=3\sin^2\theta$
$\Rightarrow5\cos^2\theta=3$
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Question 351 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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MCQ 361 Mark
The direction ratios of the diagonal of the cube joining the origin to the opposite corner are $($when the $3$ concurrent edges of the cube are coordinate axes$).$
  • A
    $\frac{2}{\sqrt{3}},\frac{2}{3},\frac{2}{3}$
  • ✓
    $1, 1, 1$
  • C
    $2, −2, 1$
  • D
    $1, 2, 3$
Answer
Correct option: B.
$1, 1, 1$
Since, a cube is a symmetric figure, the vertex we are talking about will be at the diagonally opposite end of the origin. i.e. it will be equally inclined to the three axes.
Let the side of the cube be a, then the corner opposite to origin will have coordinates $(a, a, a).$
Direction ratios of a line joining two points $(x_1​, y_1​, z_1​)$ and $(x_2​, y_2​, z_2​)$ is given by $(x_{2​ }− x_1​, y​_{2 }− y_1​, z_{2​ }− z_1​)$
Then, direction ratios of two point $(0, 0, 0)$ and $(a, a, a)$ will be $(a − 0, a − 0, a − 0) = (a, a, a) = a(1, 1, 1)$
Hence, the direction ratios are $1, 1, 1.$
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Question 371 Mark
If a line has direction ratios 2, -1, -2, determine its direction cosines:
  1. $\frac{1}{3}, \frac{2}{3},\frac{-1}{3}$
  2. $\frac{2}{3}, \frac{-1}{3},\frac{-2}{3}$
  3. $\frac{-2}{3}, \frac{1}{3}, \frac{2}{3}$
  4. None of the above
Answer
  1. $\frac{2}{3}, \frac{-1}{3},\frac{-2}{3}$
Solution:
Direction cosines are.
$=\frac{2}{2^2+(-1)^2+(-2)^2},\frac{1}{2^2+(-1)^2+(-2)^2},\frac{-2}{2^2+(-1)^2+(-2)^2}$
$=\frac{2}{3}, \frac{-1}{3},\frac{-2}{3}$
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Question 381 Mark
The line x = 1, y = 2 is:
  1. Parallel to x-axis
  2. Parallel to y-axis
  3. Parallel to z-axis
  4. None of these
Answer
  1. Parallel to z-axis
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Question 391 Mark
The distance between the point (3, 4, 5) and the point where the line $\frac{\text{x}-3}{\text{1}}=\frac{\text{y}-4}{\text{2}}=\frac{\text{z}-5}{\text{2}}$ meets the plane x + y + z = 17 is:
  1. 1
  2. 2
  3. 3
  4. None of these
Answer
  1. 3
Solution:
The coordinates of any point on the given line are of the from
$\frac{\text{x}-3}{\text{1}}=\frac{\text{y}-4}{\text{2}}=\frac{\text{z}-5}{\text{2}}=\lambda$
$\Rightarrow \text{x}=\lambda+3;\text{y}=2\lambda+4;\text{z}=2\lambda+5$
So, the coordinates of the point on the given line are $(\lambda+3,2\lambda+4,2\lambda+5)$
This point lies on the plane
x + y + z = 17
$\Rightarrow\lambda+3,2\lambda+4+2\lambda+5=17$
$\Rightarrow5\lambda=5$
$\Rightarrow\lambda=1$
So, the coordinates of the point are
$(\lambda+3,2\lambda+4,2\lambda+5)$
$=(1+3,2(1))+4,2(1)+5)$
$=(4,6,7)$
Now, the distance between the points (4, 6, 7) and (3, 4, 5) is
$\sqrt{(3+4)^2+(4-6)^2+(5-7)^2}$
$\sqrt{1+4+4}$
$=3\text{ units}$
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Question 401 Mark
A(3, 2, 0), B(5, 3, 2) and  C(-9, 6, -3) are the vertices of a tringle ABC. if the bisector of $\angle\text{ABC}$ meets BC at D, then coordinates of D are:
  1. $\Big(\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
  2. $\Big(-\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
  3. $\Big(\frac{19}{8},-\frac{57}{16},\frac{17}{16}\Big)$
  4. $\text{none of these}$
Answer
  1. $\Big(\frac{19}{8},\frac{57}{16},\frac{17}{16}\Big)$
Solution:
Since the bisector of $\angle\text{ABC}$ cannot meet BC, the solution of this quation is not possible.
Disclaimer: This quation is wrong, so the solution has not been provide.
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Question 411 Mark
The distance of the point (-3, 4, 5) from the origin:
  1. 50
  2. $5\sqrt{2}$
  3. 6
  4. None of these
Answer
  1. $5\sqrt{2}$
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Question 421 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 431 Mark
Ratio in which the xy-plane divided the join of (1, 2, 3) and (4, 2, 1) is:
  1. 3 : 1 internally
  2. 3 : 1 externally
  3. 2 : 1 internally
  4. 2 : 1 externally
Answer
  1. 3 : 1 externally
Solution:
Suppose the XY-plane divides the line segment joining the points P(1, 2, 3) and Q(4, 2, 1) in the ratio k : 1.
Using the section formula, the coordinates of the point of intersection are given by
$\Big(\frac{\text{k}(4)+1}{\text{k}+1},\frac{\text{k}(2)+2}{\text{k}+1},\frac{\text{k}(1)+3}{\text{k}+1}\Big)$
The Z-coordinate of any point on the XY-plane is zero
$\Rightarrow\frac{\text{k}(1)+3}{\text{k}+1}=0$
$\Rightarrow\text{k}+3=0$
$\Rightarrow\text{k}=-3=\frac{-3}{1}$
Thus, the XY-plane divided the line segment joining the given points in the ratio 3 : 1 externally.
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Question 441 Mark
A normal to the plane x = 2 is:
  1. < br > (0, 1, 1) < br >
  2. < br > (2, 0, 2) < br >
  3. < br > (1, 0, 0) < br >
  4. < br > (0, 1, 0) < br >
Answer
  1. < br > (0, 1, 1) < br >
Solution:
The plane x = 2 is perpendicular to x axis So the angle is $\frac{\pi}{2},\cos\frac{\pi}{2}=0$
0 The plane x = 2 is parallel to both y axis and z axis So the angle is (0, 1, 1)
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Question 451 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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MCQ 461 Mark
Which of the following triplets give the direction cosines of a line$:$
  • A
    $1, 1, 1$
  • B
    $1, -1, 1$
  • C
    $1, 1, -1$
  • ✓
    $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
Answer
Correct option: D.
$\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
If $l, m, n$ are the directions cosine of a line then $l^2+ m^2 + n^2 = 1$ Thus we get $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
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Question 471 Mark
The projections of a line segment on x, y and z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are:
  1. $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$
  2. $19;\frac{12}{19},\frac{4}{19},\frac{3}{19}$
  3. $11;\frac{12}{11},\frac{14}{11},\frac{3}{11}$
  4. $\text{None of these}$
Answer
  1. $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$
Solution:
If a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\dots(1)$
Let r be the length of the line segment. then,
$\text{r}\cos\alpha=12,\text{r}\cos\beta=4,+\cos\gamma=3\dots(2)$
$\Rightarrow\big(\text{r}\cos\alpha\big)^2+\big(\text{r}\cos\beta\big)^2+\big(\text{r}\cos\gamma\big)^2=12^2+4^3+3^2$
$\Rightarrow\text{r}^2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)=169$
$\Rightarrow\text{r}^2(1)=169$ [From (1)]
$\Rightarrow\text{r}=\sqrt{169}$
$\Rightarrow\text{r}=\pm13$
$\Rightarrow\text{r}=13$ (Since length cannot be negative)
(Since legth cannot be negative)
Substituting r = 13 in (2), we get
$\cos\alpha=\frac{12}{13},\cos\beta\frac{4}{13},\cos\gamma=\frac{1}{13}$ Thus, the direction cosines of the line are $\frac{12}{13},\frac{4}{13},\frac{1}{13}$
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Question 481 Mark
The image of the point (1, 3, 4) in the plane 2x - y + z + 3 = 0 is:
  1. (3, 5, 2)
  2. (-3, 5, 2)
  3. (3, 5, -2)
  4. (3, -5, 2)
Answer
  1. (-3, 5, 2)
Solution:
Let Q be the image of the point P(1, 3, 4) in the plane 2x - y +z + 3 = 0
Then PQ is normal to the plane.
So, the direction ratios of PQ are proportional to 2, -1, 1 equation of PQ is
Let the coordinates of Q be (2r + 1, -r + 3, r + 4)
Let R be the mid point of PQ.
Then,
$\text{R}=\Big(\frac{2\text{r}+1+1}{2},\frac{-\text{r}+3+3}{2},\frac{\text{r}+4+4}{2}\Big)$
$=\Big(\text{r}+1,\frac{-\text{r}+6}{2},\frac{\text{r}+8}{2}\Big)$
Since R lies in the plane 2x - y + z + 3 = 0,
$2(\text{r}+1)-\Big(\frac{-\text{r}+6}{2}\Big)+\frac{\text{r}+8}{2}+3=0$
⇒ 4r + 4 + r - 6 + r + 8 + 6 = 0
⇒ 6r + 12 = 0
⇒ r = -2
Substituting this in the coordinates of Q, we get
Q = (2r + 1, -r + 3, r + 4)
=(2 (-2) + 1, 2 + 3, -2 + 4)
=(-3, 5, 2).
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Question 491 Mark
If points (1, 2), (3, 5) and (0, b) are collinear the value of b is:
  1. $\frac{1}{2}$
  2. $\frac{7}{2}$
  3. $2$
  4. $-1$
Answer
  1. $\frac{1}{2}$
Solution:
Area $=\frac{1}{2}|1(5-\text{b})+3(\text{b}-2)+0(2-5)|$
As points are collinear, so area = 0
$\therefore\frac{1}{2}|1(5-\text{b})+3(\text{b}-2)+0(2-\text{5})|=0$
⇒ 5 − b + 3b − 6 = 0
⇒ = 1 = 2b
$\therefore\text{b}=\frac{1}{2}$
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Question 501 Mark
The angle between the straight lines $\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$ and $\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$ is:
  1. 45°
  2. 30°
  3. 60°
  4. 90°
Answer
  1. 90°
Solution:
We have
$\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$
$\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}-3}{-3}$
The direction of the given lines are propotional to 2, 5, 4 are 1, 2, -3.
The given lines are parallel to the vectors $\vec{\text{b}}_1=2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}.$
Let $\theta$ be the angle between the given lines.
Now,
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{2^2+5^2+4^2}\sqrt{1^2+2^2+(-3)^2}}$
$=\frac{2+10-12}{\sqrt{45}\sqrt{14}}$
$\Rightarrow\theta=90^\circ$
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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip