Maths M.C.Q (1 Marks)

3. $\frac{1}{6}$

Solution:

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 = d₁ and ax + by + cz = d₂ is,

$=\frac{|\text{d}_2-\text{d}_1|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$

So, the required distance

$=\frac{|-5+4|}{\sqrt{4^2+4^2+(-2)^2}}$

$=\frac{|-1|}{\sqrt{16+16+4}}$

$=\frac{1}{\sqrt{36}}$

$=\frac{1}{6}\text{units}$

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.

$\lambda(\text{x}-2)+(-4\lambda)(\text{y}-3)+2\lambda(\text{z}-3)=0$

$\Rightarrow\text{x}-4\text{y}+2\text{z}+4=0$

3. 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

$\sqrt{(3+4)^2+(4-6)^2+(5-7)^2}$

$\sqrt{1+4+4}$

$=3\text{ units}$

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

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

$\text{P}=\frac{\Big|\big(2\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big),\big(\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\big)-5\Big|}{\Big|\hat{\text{i}}+5\hat{\text{j}}+\hat{\text{k}}\Big|}$

$=\frac{|2-10+3-5|}{\sqrt{1+25+1}}$

$=\frac{|-10|}{\sqrt{27}}$

$=\frac{10}{3\sqrt{3}}\text{units}$

1. 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.

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

2. 60°

Solution:

We know that the angle between the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 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, $\cos\theta=\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}}$

$=\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}$

$=\frac{3}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}$

$\Rightarrow\theta\cos^{-1}\Big(\frac{1}{2}\Big)=60^\circ$

1. x - 5y + 3z = 7

Solution:

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 a₁a₂ + b₁b₂ + c₁c₂ = 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

$\Big(1+2\Big(\frac{-2}{3}\Big)\Big)\text{x}+\Big(1-\Big(\frac{-2}{3}\Big)\Big)\text{y}+\Big(1+3\Big(\frac{-2}{3}\Big)\Big)\text{z}+3+\Big(\frac{-2}{3}\Big)=0$

⇒ -x + 5y - 3z + 7 = 0

⇒ x - 5y + 3z = 7

3. (-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).

3. 51x - 15y - 50z + 173 = 0

Solution:

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 a₁a₂ + b₁b₂ + c₁c₂ = 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

$\Big(1+2\Big(\frac{-29}{7}\Big)\Big)\text{x}+\Big(2+\Big(\frac{-29}{7}\Big)\Big)\text{y}+\Big(3+\frac{29}{7}\Big)\text{z}-4+5\Big(\frac{-29}{7}\Big)=0$

⇒ 51x + 15y - 50z + 173 = 0.

3. $2\hat{\text{i}}+7\hat{\text{j}}-13\hat{\text{k}}$

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.

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

3. $\frac{7}{5}$

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

$=\frac{|3(0)+0+4(0)-7|}{\sqrt{3^2+0^2+4^2}}$

$=\frac{|0+0-7|}{\sqrt{25}}$

$=\frac{7}{5}\text{ units}$

1. (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)

4. 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.

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

$\vec{\text{r}}.\vec{\text{n}}=\vec{\text{a}}.\vec{\text{n}}$

$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(\hat{\text{i}}-\hat{\text{j}}-0\hat{\text{k}}\big).\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)$

$\Rightarrow\vec{\text{r}}.\big(5\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)=5+2+0$

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

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

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

$\frac{\text{x}}{\alpha}+\frac{\text{y}}{\beta}+\frac{\text{z}}{\gamma}=1$

$\Rightarrow\frac{\text{x}}{3\alpha}+\frac{\text{y}}{3\beta}+\frac{\text{z}}{3\gamma}=1\ [\text{From (1)}]$

$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=3$

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)

$=\sqrt{(1-2)^2+(-5+1)^2+(-10-2)^2}$

$=\sqrt{9+16+144}$

$=13 \text{ units}$