Maths M.C.Q (1 Marks)

4. $\sqrt{\alpha^2+\gamma^2}$

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

$=\big|\overrightarrow{\text{AB}}\times\overrightarrow{\text{BC}}\big|=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\2&-1&-2\\2&2&1 \end{vmatrix}$

$=|3\vec{\text{i}}-6\vec{\text{j}}+6\vec{\text{k}}|$

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

$=9\text{ sq.units}$

3. -1

Solution:

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

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

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

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

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

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.

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. 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. $\frac{2}{7}$

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, $\sin\theta=\frac{|\vec{\text{b}}\cdot\vec{\text{n}}|}{|\vec{\text{b}|}\cdot|\vec{\text{n}}|}$

$=\frac{\big|(\vec{\text{i}})\cdot(2\vec{\text{i}}-3\vec{\text{j}}+6\vec{\text{k}})\big|}{\sqrt{1}\sqrt{4+9+36}}$

$=\frac{2}{7}$

3. (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.

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)

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

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

$=13 \text{ units}$

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)]

$=2(1)-3$

$=-1$

1. $\underline{+}\frac{1}{59}$

Solution:

Give, a, 3a, 7a  be the direction cosines of a directed line.

Then from the property of direction cosines we get 

a² + (3a)² + (7a)² = 1 or 

59a² = 1 or

$\text{a}=\underline{+}\frac{1}{\sqrt{59}}$

2. 2

Solution:

For a vector.

$\cos^2(\alpha)+\cos^2(\beta)+\cos^2(\gamma)=1$

$1-\sin^2(\alpha)+1-\sin^2(\beta)+1-\sin^2(\gamma)=1$

$\sin^2(\alpha)\sin^2(\beta)+\sin^2(\gamma)=3-1=2$

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

Solution:

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

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

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

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

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

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

4. 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.

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

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 >

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)

$=\sqrt{(\text{a}-\text{a})^2+(\text{b}-0)^2+(\text{c}-0)^2}$

$=\sqrt{\text{b}^2+\text{c}^2}$

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. $\frac{1}{3}$

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 =

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

$\text{n}=\frac{-2}{1^2+2^2+(-2)^2}=\frac{-2}{3}$

$\Rightarrow\text{l + m + n}=\frac{1}{3}$

1. 7

Solution:

$\frac{\text{x}-6}{3}=\frac{\text{y}-7}{2}=\frac{\text{z}-7}{-2}$

Let point (1, 2, 3) be P and the point through which the line passes be Q(6, 7,  7). Also, the line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$

Now,

$\overrightarrow{\text{PQ}}=5\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$

$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}} =\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&2&-2\\5&5&4\end{vmatrix}$

$=18\hat{\text{i}}-22\hat{\text{j}}+5\hat{\text{k}}$

$\Rightarrow\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{18^2+(-22)^2+5^2}$

$=\sqrt{324+484+25}$

$=\sqrt{833}$

$\because\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$

$=\frac{\sqrt{833}}{\sqrt{17}}$

$=\sqrt{49}$

$=7$