Maths M.C.Q (1 Marks)

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.

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,

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

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

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

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,

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

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

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

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

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

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

1. $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$

Solution:

If a line makes angles $\alpha,\beta$ and $\gamma$ with the axes, then $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\dots(1)$

Let r be the length of the line segment. then,

$\text{r}\cos\alpha=12,\text{r}\cos\beta=4,+\cos\gamma=3\dots(2)$

$\Rightarrow\big(\text{r}\cos\alpha\big)^2+\big(\text{r}\cos\beta\big)^2+\big(\text{r}\cos\gamma\big)^2=12^2+4^3+3^2$

$\Rightarrow\text{r}^2\big(\cos^2\alpha+\cos^2\beta+\cos^2\gamma\big)=169$

$\Rightarrow\text{r}^2(1)=169$ [From (1)]

$\Rightarrow\text{r}=\sqrt{169}$

$\Rightarrow\text{r}=\pm13$

$\Rightarrow\text{r}=13$ (Since length cannot be negative)

(Since legth cannot be negative)

Substituting r = 13 in (2), we get

$\cos\alpha=\frac{12}{13},\cos\beta\frac{4}{13},\cos\gamma=\frac{1}{13}$ Thus, the direction cosines of the line are $\frac{12}{13},\frac{4}{13},\frac{1}{13}$

4. 90°

Solution:

We have

$\frac{\text{x}+1}{2}=\frac{\text{y}-2}{5}=\frac{\text{z}+3}{4}$

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

The direction of the given lines are propotional to 2, 5, 4 are 1, 2, -3.

The given lines are parallel to the vectors $\vec{\text{b}}_1=2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}.$

Let $\theta$ be the angle between the given lines.

Now,

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

$=\frac{\big(2\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\big).\big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)}{\sqrt{2^2+5^2+4^2}\sqrt{1^2+2^2+(-3)^2}}$

$=\frac{2+10-12}{\sqrt{45}\sqrt{14}}$

$\Rightarrow\theta=90^\circ$

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. 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,

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

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

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

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

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

$=\sqrt{833}$

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

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

$=\sqrt{49}$

$=7$

3. $\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

Solution:

Equation of the line passing through the points having position vectors

$\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:

$\vec{r}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big\{\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)-\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)\big\},$ where t is a parameter

$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

$=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$

1. $3,1,-2$

Solution:

We have

x - y + z - 5 = 0 = x - 3y - 6

⇒ x - 3y - 6=0

x - y + z - 5 = 0

⇒ x = 3y + 6 .....(1)

x - y + z - 5 = 0.....(2)

From (1) and (2), we get

3y + 6 - y + z - 5 = 0

⇒ 2y + z + 1 = 0

$\Rightarrow\text{y}=\frac{-\text{z}-1}{2}$

$\text{y}=\frac{\text{x}-6}{3}$ [From (1)]

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

So, the given equation can be re-witten as

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

Hence, the direction ratios the given line are proportional to 3, 1, -2.

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

Solution:

If a line makes angles $\alpha,\beta$ and $\gamma$ with the axcs, then $\cos2\alpha+\cos2\beta+\cos2\gamma=1.$

Here,

$\alpha=\frac{\pi}{3}$

$\beta=\frac{\pi}{4}$

Now,

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

$\Rightarrow\cos^2\frac{\pi}{3}+\cos^2\frac{\pi}{4}+\cos^2\gamma=1$

$\Rightarrow\frac{1}{4}+\frac{1}{2}+\cos^2\gamma=1 $

$\Rightarrow\cos^2\gamma=1-\frac{3}{4}$

$\Rightarrow\cos^2\gamma=\frac{1}{4}$

$\Rightarrow\cos\gamma=\frac{1}{2}$

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

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. 4, 5, 7

Solution:

We have

$\frac{\text{x}-7}{2}=\frac{\text{y}+17}{-3}=\frac{\text{z}-6}{1}$

$\frac{\text{x}+5}{1}=\frac{\text{y}+3}{2}=\frac{\text{z}-4}{-2}$

The direction ratios of the given lines are proportional to 2, -3, 1 and 1, 2, -2.

The vectors parallel to the given vectors are $\vec{\text{b}}_1=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}_2=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$

Vector perpendicular to the given two lines is

$\vec{\text{b}}=\vec{\text{b}}_1\times\vec{\text{b}}_2$

$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&1\\1&2&-2\end{vmatrix}$

$=4\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$

Hence, the direction ration of the line perpendicular to the given two lines are proportional to 4, 5, 7.

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

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

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