Maths M.C.Q (1 Marks)

3. -1

Solution:

$\cos^2\alpha+\cos^2\beta+\cos^2\text{r}=1\cos2\alpha+\cos^2\beta+\cos2\text{r}$

$=2\cos^2\alpha-1+2\cos^2\beta-1+2\cos^2\text{r}-1$

$=2(\cos^2\alpha+\cos^2\beta\cos{\text{r}})-3=2(1)-3=-1$

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

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

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

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

4. 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

$\text{PABC}=\frac{125}{8}\text{cubic units}$

3. 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

2. $4$

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 $=4\text{k}+2$

$=4\Big(\frac{1}{2}\Big)+2$

$=2+2=4$

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, $\cos\theta=\frac{(2)(1)+(-1)(1)+(1)(2)}{\sqrt{2^2+(-1)^2+1^2}\sqrt{1^2+1^2+2^2}}$

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

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

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

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. -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)]

$=2(1)-3$

$=-1$

3. $\text{c}=\underline{+}\sqrt{2}$

Solution:

Since, DC′s of a line are $\Big(\frac{1}{\text{c}},\frac{1}{\text{c}},\frac{1}{\text{c}}\Big)$

$\because\text{l}^2 + \text{m}^2 + \text{n}^2 = 1$

$\because\Big(\frac{1}{\text{c}}\Big)^2+\Big(\frac{1}{\text{c}}\Big)^2+\Big(\frac{1}{\text{c}}\Big)^2=1$

$\Rightarrow1 + 1 + 1 = \text{c}^2$

$\Rightarrow\text{c}^2 = 3$

$\Rightarrow\text{c}=\underline{+}\sqrt{3}$

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)

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

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

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

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

2. 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

$=\Big(\frac{1}{\sqrt{3}}\Big)^2+\Big(\frac{2}{\sqrt{3}}\Big)^2+\Big(\frac{-2}{\sqrt{3}}\Big)^2=\frac{1+4+4}{3}=3$

3. $\frac{\pi}{3}$

4. none of these

Solution:

Suppose, l, m, n are direction cosines

⟹ 1² + m² + n² = 1

But 1 = m = n = 1

⟹ 3m² = 1

⟹ 1 = m = n = $\frac{1}{\sqrt3}$​

which are not direction cosines of either of the three co-ordinate axes.

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.

4. perpendicular to z-axis

Solution:

We have

$\frac{\text{x}-3}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-1}{0}$

Also, the given line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$

Let $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ be parpendicular to the given line.

Now,

$3\text{x}+4\text{y}+0\text{z}=0$

It is satisfied by the coordinates of z-axis, i.e. (0, 0, 1).

Hence, the given line is perpendicular to z-axis.

4. $(\alpha,\beta,\gamma)$

4. $3\sqrt{30}$

Solution:

We have

$\frac{\text{x}-3}{3}=\frac{\text{y}-8}{-1}=\frac{\text{z}-3}{1}\dots(1)$

$\frac{\text{x}+3}{-3}=\frac{\text{y}+7}{2}=\frac{\text{z}-6}{4}\dots(2)$

We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3, -1, 1.

Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1,$ where $\vec{\text{a}}_1=3\hat{\text{i}}+8\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}_1=3\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$

Also, line (2) passes through the point (3, -7, 6) and has direction ratios proprtional to -3, 2, 4.

Its vector equation is $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2,$ where $\vec{\text{a}}_2=-3\hat{\text{i}}-7\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}_2=-3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}.$

Now,

$\vec{\text{a}}_2-\vec{\text{a}}_1=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$

$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-1&1\\-3&2&4\end{vmatrix}$

$=-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}$

$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-6)^2+(-15)^2+3^2}$

$=\sqrt{36+225+9}$

$=\sqrt{270}$

$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big).\big(-6\hat{\text{i}}-15\hat{\text{j}}+3\hat{\text{k}}\big)$

$=36+225+9$

$=270$

The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by

$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$

$=\Big|\frac{270}{\sqrt{270}}\Big|$

$=\sqrt{270}$

$=3\sqrt{30}$

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

$\frac{1}{\sqrt{1^2+(-3)^2+2^2}},\frac{-3}{\sqrt{1^2+(-3)^2+2^2}},\frac{2}{\sqrt{1^2+(-3)^2+2^2}}$

$=\frac{1}{\sqrt{14}},-\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}$

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

4. $\frac{40}{49}$

1. $\frac{9}{11},\frac{6}{11},\frac{2}{11}$

Solution:

Dr's of the line are : 18, 12, 4

$\text{Dc's}=\frac{18}{\sqrt{18^2+12^2+4^2}},\frac{12}{\sqrt{18^2+12^2+4^2}},\frac{4}{\sqrt{18^2+12^2+4^2}}$

$=\frac{18}{22},\frac{12}{22},\frac{4}{22}$

$=\frac{9}{11},\frac{6}{11},\frac{2}{11}$