Maths M.C.Q (1 Marks)

2. Non collinear

Solution:

Area of triangle formed by these vertices is,

$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\2&3&1\\8&11&1\end{vmatrix}$

Applying R₂​ → R₂​ − R₁​, R_3​ → R_{3 ​}− R1​

$\triangle=\frac{1}{2}\begin{vmatrix}1&1&1\\1&2&0\\7&10&0\end{vmatrix}=\frac{1}{2}(10-14)=2$

Hence points are non collinear

2. $\underline{+}\frac{1}{\sqrt{3}}$

3. -1

Solution:

Suppose the point P divided the line segment joining the point Q(2, 2, 1) and R(5, 1, -2) in the ratio k : 1.

Using the section formula, the coordinates of the point of intersection are given by

$\Big(\frac{\text{k}(5)+2}{\text{k}+2},\frac{\text{k}(1)+2}{\text{k}+1},\frac{\text{k}(-2)+1}{\text{k}+1}\Big) $

On the XY-plane, the Z-coordinate of any point is zero.

$\Rightarrow\frac{\text{k}(5)+2}{\text{k}+2}=4$

$\Rightarrow5\text{k}+2=4(\text{k}+1)$

$\Rightarrow\text{k}=2$

Now,

Z-coordinate of P $=\frac{\text{k}(-2)+1}{\text{k}+1}$ 

$\frac{2(-2)+1}{2+1}\ [\text{Substituting} \text{ k}=2]$

$=-1$

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

Solution:

In XY-plane, only the sign of z coordinate of the point got changed after the reflection. Therefore, the reflection of the point $(\alpha,\beta,\gamma)$ is $(\alpha,\beta,-\gamma).$

2. 1

Solution:

As the points are collinear, the slope of the line joining

any two points, should be same as the slope of the line joining two other points.

Slope of the line passing through points (x₁​, y₁​) and $\text{x}_2,\text{y}_2=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}$

So, slope of the line joining (p, 0), (0, q) = Slope of the line joining

(0, q) and (1, 1)

$\frac{\text{q}-0}{0-\text{p}}=\frac{1-\text{q}}{1-\text{p}}$

$-\frac{\text{q}}{\text{p}}=1-\text{q}$

Dividing both sides by q,

$-\frac{1}{\text{p}}=\frac{1}{\text{q}}-1$

$\Rightarrow\frac{1}{\text{p}}+\frac{1}{\text{q}}=1$

1. Coincident.

Solution:

The equation of the given lines are

$\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}\dots(1)$

$\frac{\text{x}-1}{-2}=\frac{\text{y}-2}{-4}=\frac{\text{z}-3}{-6}$

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

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,

$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)\times\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$

$=\vec{0}$ $\big[\because\vec{\text{a}}\times\vec{\text{a}}=\vec{0}\big]$

2. 3 : 2 externally

Solution:

Suppose the point R divides PQ in the ratio $\lambda:1$.

Coordinates of R are $\Big(\frac{5\lambda+3}{\lambda+1},\frac{4\lambda+2}{\lambda+1},\frac{-6\lambda-4}{\lambda+1}\Big)$.

But the coordinates of R are (9, 8, -10).

$\therefore\frac{5\lambda+3}{\lambda+1}=9,\frac{4\lambda+2}{\lambda+1}=8$ and $\frac{-6\lambda-4}{\lambda+1}=-10$

From each of these equations, we get

$\lambda=-\frac{3}{2}$

$\therefore$ R divided PQ in the ratio 3 : 2 externally.

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

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,

$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$

$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big\{\big(-\sqrt{3}-1\big)\hat{\text{i}}+\big(\sqrt{3}-1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}}{\sqrt{1^2+1^2+1^2}\sqrt{\big(-\sqrt{3}-1\big)^2+\big(\sqrt{3}-1\big)^2+4^2}}$

$=\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{3}\sqrt{24}}$

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

$\frac{1}{\sqrt{2}}$

$\Rightarrow\theta=\frac{\pi}{3}$

4. $2$

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

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

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

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

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.

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

1. 1.

Solution:

The general equation of a plane in vector form is given by $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$

Where d is the distance of the plane from the origin.

Comparing $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$ and $\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=1,$ we get

Therefore, the distance of the plane $\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=1$ from the origin is 1.

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

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

Solution:

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

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

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

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

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

1. 5x + y − 11 = 0

Solution:

Since, the points A(1, 6), B(3, −4) and C(x, y) are colinear

$\therefore\begin{vmatrix}1&6&1\\3&-4&1\\\text{x}&\text{y}&1\end{vmatrix}=0$

⇒ 1(−4−y) −6(3 − x) + 1(3y + 4x) = 0

⇒ 10x + 2y − 22 = 0

⇒ 5x + y − 11 = 0

4. $\sqrt{\text{a}^2+\gamma^2}$

Solution:

Required distance $=\sqrt{(\alpha-0)^2+(\beta-\beta)^2+(\gamma-0)^2}$

$=\sqrt{\alpha+\gamma^2}$

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

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

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

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

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

3. $1, 2, 3$

Solution:

6x - 2 = 3y + 1 = 2x - 2

$6\big(\text{x}-\frac{2}{6}\big)=3(\text{y}+\frac{1}{3}\big)=2(\text{x}-\frac{2}{2}\big)$

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

Line will be passing through the poits, $\Big(\frac{1}{3},-\frac{1}{3},1\Big)$

and parallel to the line having direction ratios is 1, 2, 3

1. 30°

Solution:

Given equation

$\text{r}[\cos\theta−3​\sin\theta]=5 $

$\text{x}−\sqrt{3}\text{​y}=5$

Slope of the line is $\tan\theta=\frac{1}{\sqrt{3}}​$

$\Rightarrow\theta=30^\circ$

Hence, the line makes an angle of 30° with the initial line.

2. $\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$

Solution:

Given,

a = (-1, 5, 4)

b = (0, 0, 1) [$\therefore$ 1 to plone z]

We know that,

$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$

$\vec{\text{r}}=(-\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})+\lambda\hat{\text{k}}$

$\Rightarrow\vec{\text{r}}=-\hat{\text{i}}+5\hat{\text{j}}+(4+\lambda)\hat{\text{k}}$

1. $\Big(\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}},\frac{6}{\sqrt{3}}\Big)$

4. $\frac{\pi}{2}$

2. $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0$

Solution:

Equation of line bisecting XOY is x = y

$\therefore$ d.r.s are (1, 1, 0)

And thus d.c.s are $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Big)$