M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

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18 questions · 4 auto-graded MCQ + 14 self-marked written.

MCQ 11 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
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 21 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:

x - 4y + 2z + 4 = 0
x + 4y + 2z + 4 = 0
x - 4y + 2z - 4 = 0
None of these

Answer

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 31 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
2
3
None of these

Answer

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 41 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:

$\frac{5}{3\sqrt{3}}$
$\frac{10}{3\sqrt{3}}$
$\frac{25}{3\sqrt{3}}$
$\text{None of these}$

Answer

$\frac{10}{3\sqrt{3}}$

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 51 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 61 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:

$\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$
$\vec{\text{r}}.(\hat{\text{i}}-3\hat{\text{k}})=10$
$\vec{\text{r}}.(3\hat{\text{i}}+\hat{\text{k}})=10$
$\text{None of these}$

Answer

$\vec{\text{r}}.(\hat{\text{i}}+3\hat{\text{k}})=10$

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 71 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 81 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 91 Mark

The image of the point (1, 3, 4) in the plane 2x - y + z + 3 = 0 is:

(3, 5, 2)
(-3, 5, 2)
(3, 5, -2)
(3, -5, 2)

Answer

(-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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MCQ 101 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 111 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:

$-2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$
$2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$
$-2\hat{\text{i}}-7\hat{\text{j}}+13\hat{\text{k}}$
$2\hat{\text{i}}+7\hat{\text{j}}+13\hat{\text{k}}$

Answer

$2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$

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 121 Mark

The eqution of the plane which cute equal intercepts of unit length on the coordinate axes is:

x + y + z = 1
x + y + z = 0
x + y - z = 1
x + y + z = 2

Answer

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 131 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:

$\frac{3}{4}$
$\frac{4}{3}$
$\frac{7}{5}$
$1$

Answer

$\frac{7}{5}$

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 141 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 151 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 161 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:

$\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$
$\vec{\text{r}}.(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=7$
$\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})=7$
$\text{None of these}$

Answer

$\vec{\text{r}}.(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}})=7$

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 171 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
2
3
None of these

Answer

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 181 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:

9
13
17
None of these

Answer

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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M.C.Q (1 Marks) - MATHS STD 12 Science Questions - Vidyadip