NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

Question 1: If a line makes angles $90^{\circ}, 135^{\circ},45^{\circ}$ with the x, y and z-axes respectively, find its direction cosines.

Answer:

Let the direction cosines of the line be l, m, and n.

So, we have

$l = \cos90^{\circ}=0$

$m = \cos135^{\circ}=-\frac{1}{\sqrt2}$

$n= \cos45^{\circ}=\frac{1}{\sqrt2}$

Therefore, the direction cosines of the lines are $0,\ -\frac{1}{\sqrt2}, and\ \ \frac{1}{\sqrt2}$.

Question 2: Find the direction cosines of a line which makes equal angles with the coordinate axes.

Answer:

If the line is making an equal angle with the coordinate axes. Then,

Let the common angle made be $\alpha$ with each coordinate axis.

Therefore, we can write;

$l = \cos \alpha,m= \cos \alpha, and\ n= \cos \alpha$

And as we know the relation, $l^2+m^2+n^2 = 1$

$\Rightarrow \cos^2 \alpha +\cos^2 \alpha+\cos^2 \alpha = 1$

$\Rightarrow \cos^2 \alpha = \frac{1}{3}$

or $\cos \alpha =\pm \frac{1}{\sqrt3}$

Thus the direction cosines of the line are $\pm \frac{1}{\sqrt3},\ \pm \frac{1}{\sqrt3},and\ \pm \frac{1}{\sqrt3}$

Question 3: If a line has the direction ratios –18, 12, – 4, then what are its direction cosines?

Answer:

Given a line has direction ratios of -18, 12, – 4, then its direction cosines are;

A line having direction ratio -18 has direction cosine:

$\frac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}} = \frac{-18}{22} = \frac{-9}{11}$

A line having direction ratio 12 has direction cosine:

$\frac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}} = \frac{12}{22} =\frac{6}{11}$

A line having direction ratio -4 has direction cosine:

$\frac{12}{\sqrt{(-4)^2+(12)^2+(-4)^2}} = \frac{-4}{22} = \frac{-2}{11}$

Thus, the direction cosines are $\frac{-9}{11},\ \frac{6}{11},\ \frac{-2}{11}$ .

Question 4: Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.

Answer:

We have the points, A (2, 3, 4), B (– 1, – 2, 1), C (5, 8, 7);

And as we can find the direction ratios of the line joining the points $(x_{1},y_{1},z_{1}) \ and\ (x_{2},y_{2},z_{2})$ is given by $x_{2}-x_{1}, y_{2}-y_{1}, \ and\ z_{2}-z_{1}.$

The direction ratios of AB are $(-1-2), (-2-3),\ and\ (1-4)$
i.e., $-3,\ -5,\ and\ -3$

The direction ratios of BC are $(5-(-1)), (8-(-2)),\ and\ (7-1)$
i.e., $6,\ 10,\ and\ 6$ .

We can see that the direction ratios of AB and BC are proportional to each other and are -2 times.

$\therefore$ AB is parallel to BC, and as point B is common to both AB and BC,

Hence, points A, B, and C are collinear.

Question 5: Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Answer:

Given the vertices of the triangle $\triangle ABC$ (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Finding each side direction ratios;

$\Rightarrow$ Direction ratios of side AB are $(-1-3), (1-5),\ and\ (2-(-4))$ i.e.,

$-4,-4,\ and\ 6.$

Therefore, its direction cosines values are;

$\frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}}$ $or\ \frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$

Similarly, for side BC;

$\Rightarrow$ Direction ratios of side BC are $(-5-(-1)), (-5-1),\ and\ (-2-2)$ i.e.,

$-4,-6,\ and\ -4.$

Therefore, its direction cosines values are;

$\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}$ $or\ \frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}$

$\Rightarrow$ Direction ratios of side CA are $(-5-3), (-5-5),\ and\ (-2-(-4))$ i.e.,

$-8,-10,\ and\ 2.$

Therefore, its direction cosines values are;

$\frac{-8}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{-5}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{2}{\sqrt{(-8)^2+(10)^2+(2)^2}}$ $or\ \frac{-8}{2\sqrt{42}},\frac{-10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}\ or\ \frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}$

Question 1: Show that the three lines with direction cosines

$\frac{12}{13},\frac{-3}{13},\frac{-4}{13};\frac{4}{13},\frac{12}{13},\frac{3}{13};\frac{3}{13},\frac{-4}{13},\frac{12}{13}$ are mutually perpendicular.

Answer:

Given direction cosines of the three lines;

$L_{1}\ \left ( \frac{12}{13}, \frac{-3}{13},\frac{-4}{13} \right )$ $L_{2}\ \left ( \frac{4}{13}, \frac{12}{13},\frac{3}{13} \right )$ $L_{3}\ \left ( \frac{3}{13}, \frac{-4}{13},\frac{12}{13} \right )$

And we know that two lines with direction cosines $l_{1},m_{1},n_{1}$ and $l_{2},m_{2},n_{2}$ are perpendicular to each other, if $l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0$

Hence, we will check each pair of lines:

Lines $L_{1}\ and\ L_{2}$ ;

$l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{12}{13}\times\frac{4}{13} \right ]+\left [ \frac{-3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times \frac{3}{13} \right ]$

$\Rightarrow$$ \left [ \frac{48}{169} \right ]-\left [ \frac{36}{169} \right ]-\left [ \frac{12}{169} \right ]= 0$

$\therefore$ the lines $L_{1}\ and\ L_{2}$ are perpendicular.

Lines $L_{2}\ and\ L_{3}$ ;

$l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{4}{13}\times\frac{3}{13} \right ]+\left [ \frac{12}{13}\times\frac{-4}{13} \right ]+\left [ \frac{3}{13}\times \frac{12}{13} \right ]$

$\Rightarrow$$ \left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]+\left [ \frac{36}{169} \right ]= 0$

$\therefore$ the lines $L_{2}\ and\ L_{3}$ are perpendicular.

Lines $L_{3}\ and\ L_{1}$ ;

$l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times\frac{-3}{13} \right ]+\left [ \frac{12}{13}\times \frac{-4}{13} \right ]$

$\Rightarrow$$ \left [ \frac{36}{169} \right ]+\left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]= 0$

$\therefore$ the lines $L_{3}\ and\ L_{1}$ are perpendicular.

Thus, we have all lines that are mutually perpendicular to each other.

Question 2: Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Answer:

We have given points where the line is passing through it;

Consider the line joining the points (1, – 1, 2) and (3, 4, – 2) is AB and the line joining the points (0, 3, 2) and (3, 5, 6) is CD.

So, we will find the direction ratios of the lines AB and CD;

Direction ratios of AB are $a_{1},b_{1}, c_{1}$

$(3-1),\ (4-(-1)),\ and\ (-2-2)$ or $2,\ 5,\ and\ -4$

Direction ratios of CD are $a_{2},b_{2}, c_{2}$

$(3-0),\ (5-3)),\ and\ (6-2)$ or $3,\ 2,\ and\ 4$ .

Now, lines AB and CD will be perpendicular to each other if $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =0$

$\Rightarrow$$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =\left ( 2\times3 \right ) +\left ( 5\times2 \right )+ \left ( -4\times 4 \right )$

$\Rightarrow$$ 6+10-16 = 0$

Therefore, AB and CD are perpendicular to each other.

Question 3: Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (– 1, – 2, 1), (1, 2, 5).

Answer:

We have given points where the line is passing through them;

Consider the line joining the points (4, 7, 8) and (2, 3, 4) is AB and the line joining the points (– 1, – 2, 1) and (1, 2, 5) is CD.

So, we will find the direction ratios of the lines AB and CD;

Direction ratios of AB are $a_{1},b_{1}, c_{1}$

$(2-4),\ (3-7),\ and\ (4-8)$ or $-2,\ -4,\ and\ -4$

Direction ratios of CD are $a_{2},b_{2}, c_{2}$

$(1-(-1)),\ (2-(-2)),\ and\ (5-1)$ or $2,\ 4,\ and\ 4$ .

Now, lines AB and CD will be parallel to each other if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Therefore, we have now;

$\frac{a_{1}}{a_{2}} = \frac{-2}{2}=-1$ $\frac{b_{1}}{b_{2}} = \frac{-4}{4}=-1$ $\frac{c_{1}}{c_{2}} = \frac{-4}{4}=-1$

$\therefore \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Hence, we can say that AB is parallel to CD.

Question 4: Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector $3\widehat{i}+2\widehat{j}-2\widehat{k}$.

Answer:

It is given that the line is passing through A (1, 2, 3) and is parallel to the vector $\vec{b}=3\widehat{i}+2\widehat{j}-2\widehat{k}$

We can easily find the equation of the line which passes through the point A and is parallel to the vector $\vec{b}$ by the known relation;

$\vec{r} = \vec{a} +\lambda\vec{b}$ , where $\lambda$ is a constant.

So, we have now,

$\\\mathrm{\Rightarrow\vec{r}=\widehat{i}+2\widehat{j}+3\widehat{k} + \lambda(3\widehat{i}+2\widehat{j}-2\widehat{k})}$

Thus, the required equation of the line.

Question 5: Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\widehat{i}-\widehat{j}+4\widehat{k}$ and is in the direction $\widehat{i}+2\widehat{j}-\widehat{k}$.

Answer:

Given that the line is passing through the point with position vector $2\widehat{i}-\widehat{j}+4\widehat{k}$ and is in the direction of the line $\widehat{i}+2\widehat{j}-\widehat{k}$.

And we know the equation of the line which passes through the point with the position vector $\vec{a}$ and parallel to the vector $\vec{b}$ is given by the equation,

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

$\Rightarrow\vec{r}=2\widehat{i}-\widehat{j}+4\widehat{k} + \lambda(\widehat{i}+2\widehat{j}-\widehat{k})$

So, this is the required equation of the line in the vector form.

$\vec{r} =x\widehat{i}+y\widehat{j}+z\widehat{k} = (\lambda+2)\widehat{i}+(2\lambda-1)\widehat{j}+(-\lambda+4)\widehat{k}$

Eliminating $\lambda$, from the above equation, we obtain the equation in the Cartesian form :

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

Hence, this is the required equation of the line in Cartesian form.

Question 6: Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$.

Answer:

Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ ;

The direction ratios of the line, $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ are 3,5 and 6 .

So, the required line is parallel to the above line.

Therefore, we can take the direction ratios of the required line as 3k, 5k, and 6k, where k is a non-zero constant.

And we know that the equation of line passing through the point $(x_{1},y_{1},z_{1})$ and with direction ratios a, b, c is written by: $\frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} = \frac{z-z_{1}}{c}$.

Therefore, we have the equation of the required line:

$\frac{x+2}{3k} = \frac{y-4}{5k} = \frac{z+5}{6k}$

or $\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} = k$

The required line equation.

Question 7: The cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{7}$ . Write its vector form.

Answer:

Given the Cartesian equation of the line;

$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{7}$

Here, the given line is passing through the point $(5,-4,6)$.

So, we can write the position vector of this point as;

$\vec{a} = 5\widehat{i}-4\widehat{j}+6\widehat{k}$

And the direction ratios of the line are 3, 7, and 2.

This implies that the given line is in the direction of the vector, $\vec{b} = 3\widehat{i}+7\widehat{j}+2\widehat{k}$.

Now, we can easily find the required equation of the line:

As we know that the line passing through the position vector $\vec{a}$ and in the direction of the vector $\vec{b}$ is given by the relation,

$\vec{r} = \vec{a} + \lambda \vec{b},\ \lambda \epsilon R$

So, we get the equation.

$\vec{r}=5\widehat{i}-4\widehat{j}+6\widehat{k}+\lambda(3\widehat{i}+7\widehat{j}+2\widehat{k}),\ \lambda \epsilon R$

This is the required equation of the line in the vector form.

Question 8: Find the angle between the following pairs of lines:

(i)$\overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k})$ and $\overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})$

(ii) $\overrightarrow{r}= 3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda (\widehat{i}-\widehat{j}-2\widehat{k})$ and $\overrightarrow{r}=2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})$

Answer:

(i) To find the angle A between the pair of lines $\vec{b_{1}}\ and\ \vec{b_{2}}$ we have the formula;

$\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |$

We have two lines :

$\overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k})$ and

$\overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})$

The given lines are parallel to the vectors $\vec{b_{1}}\ and\ \vec{b_{2}}$ ;

where $\vec{b_{1}}= 3\widehat{i}+2\widehat{j}+6\widehat{k}$ and $\vec{b_{2}}= \widehat{i}+2\widehat{j}+2\widehat{k}$ respectively,

Then we have

$\vec{b_{1}}.\vec{b_{2}}=(3\widehat{i}+2\widehat{j}+6\widehat{k}).(\widehat{i}+2\widehat{j}+2\widehat{k})$

$=3+4+12 = 19$

and $|\vec{b_{1}}| = \sqrt{3^2+2^2+6^2} = 7$

$|\vec{b_{2}}| = \sqrt{1^2+2^2+2^2} = 3$

Therefore, we have;

$\cos A = \left | \frac{19}{7\times3} \right | = \frac{19}{21}$

or $A = \cos^{-1} \left ( \frac{19}{21} \right )$

(ii) To find the angle A between the pair of lines $\vec{b_{1}}\ and\ \vec{b_{2}}$ we have the formula;

$\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |$

We have two lines :

$\overrightarrow{r}=3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda(\widehat{i}-\widehat{j}-2\widehat{k})$ and

$\overrightarrow{r}=2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})$

The given lines are parallel to the vectors $\vec{b_{1}}\ and\ \vec{b_{2}}$ ;

where $\vec{b_{1}}= \widehat{i}-\widehat{j}-2\widehat{k}$ and $\vec{b_{2}}= 3\widehat{i}-5\widehat{j}-4\widehat{k}$ respectively,

Then we have

$\vec{b_{1}}.\vec{b_{2}}=(\widehat{i}-\widehat{j}-2\widehat{k}).(3\widehat{i}-5\widehat{j}-4\widehat{k})$

$=3+5+8 = 16$

and $|\vec{b_{1}}| = \sqrt{1^2+(-1)^2+(-2)^2} = \sqrt{6}$

$|\vec{b_{2}}| = \sqrt{3^2+(-5)^2+(-4)^2} = \sqrt{50} = 5\sqrt2$

Therefore, we have;

$\cos A = \left | \frac{16}{\sqrt6 \times5\sqrt2} \right | = \frac{16}{10\sqrt3}$

or $A = \cos^{-1} \left ( \frac{8}{5\sqrt3} \right )$

Question 9: Find the angle between the following pair of lines:

(i) $\frac{x-2}{2}= \frac{y-1}{5}= \frac{z+3}{-3}$ and $\frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}$

(ii) $\frac{x}{2}= \frac{y}{2}=\frac{z}{1}$ and $\frac{x -5}{4}= \frac{y-2}{1}=\frac{z-3}{8}$

Answer:

(i) Given lines are;

$\frac{x-2}{2}= \frac{y-1}{5}= \frac{z+3}{-3}$ and $\frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}$

So, we have two vectors $\vec{b_{1}}\ and\ \vec{b_{2}}$, which are parallel to the pair of above lines respectively.

$\vec{b_{1}}\ =2\widehat{i}+5\widehat{j}-3\widehat{k}$ and $\vec{b_{2}}\ =-\widehat{i}+8\widehat{j}+4\widehat{k}$

To find the angle A between the pair of lines $\vec{b_{1}}\ and\ \vec{b_{2}}$ we have the formula;

$\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |$

Then we have

$\vec{b_{1}}.\vec{b_{2}}=(2\widehat{i}+5\widehat{j}-3\widehat{k}).(-\widehat{i}+8\widehat{j}+4\widehat{k})$

$=-2+40-12 = 26$

and $|\vec{b_{1}}| = \sqrt{2^2+5^2+(-3)^2} = \sqrt{38}$

$|\vec{b_{2}}| = \sqrt{(-1)^2+(8)^2+(4)^2} = \sqrt{81} = 9$

Therefore, we have;

$\cos A = \left | \frac{26}{\sqrt{38} \times9} \right | = \frac{26}{9\sqrt{38}}$

or $A = \cos^{-1} \left ( \frac{26}{9\sqrt{38}} \right )$

(ii) Given lines are;

$\frac{x}{2}= \frac{y}{2}=\frac{z}{1}$ and $\frac{x -5}{4}= \frac{y-2}{1}=\frac{z-3}{8}$

So, we have two vectors $\vec{b_{1}}\ and\ \vec{b_{2}}$ which are parallel to the pair of above lines respectively.

$\vec{b_{1}}\ =2\widehat{i}+2\widehat{j}+\widehat{k}$ and $\vec{b_{2}}\ =4\widehat{i}+\widehat{j}+8\widehat{k}$

To find the angle A between the pair of lines $\vec{b_{1}}\ and\ \vec{b_{2}}$ we have the formula;

$\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |$

Then we have

$\vec{b_{1}}.\vec{b_{2}}=(2\widehat{i}+2\widehat{j}+\widehat{k}).(4\widehat{i}+\widehat{j}+8\widehat{k})$

$=8+2+8 = 18$

and $|\vec{b_{1}}| = \sqrt{2^2+2^2+1^2} = \sqrt{9} = 3$

$|\vec{b_{2}}| = \sqrt{(4)^2+(1)^2+(8)^2} = \sqrt{81} = 9$

Therefore, we have;

$\cos A = \left | \frac{18}{ 3\times9} \right | = \frac{2}{3}$

or $A = \cos^{-1} \left ( \frac{2}{3} \right )$

Question 10: Find the values of p so that the lines $\frac{1-x}{3}=\frac{7y-14}{2p}= \frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}= \frac{6-z}{5}$ are at right angles.

Answer:

First, we have to write the given equation of lines in the standard form;

$\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}= \frac{z-3}{2}$ and $\frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}= \frac{z-6}{-5}$

Then we have the direction ratios of the above lines as;

$-3,\ \frac{2p}{7},\ 2$ and $\frac{-3p}{7},\ 1,\ -5$ respectively..

Two lines with direction ratios $a_{1},b_{1},c_{1}$ and $a_{2},b_{2},c_{2}$ are perpendicular to each other if, $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0$

$\therefore (-3).\left ( \frac{-3p}{7} \right )+(\frac{2p}{7}).(1) + 2.(-5) = 0$

$\Rightarrow \frac{9p}{7}+ \frac{2p}{7} =10$

$\Rightarrow 11p =70$

$\Rightarrow p =\frac{70}{11}$

Thus, the value of p is $\frac{70}{11}$.

Question 11: Show that the lines $\frac{x-5}{7}=\frac{y+2}{-3}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other.

Answer:

First, we have to write the given equation of lines in the standard form;

$\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$

Then we have the direction ratios of the above lines as;

$7,\ -5,\ 1$ and $1,\ 2,\ 3$ respectively..

Two lines with direction ratios $a_{1},b_{1},c_{1}$ and $a_{2},b_{2},c_{2}$ are perpendicular to each other if, $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0$

$\therefore 7(1) + (-5)(2)+1(3) = 7-10+3 = 0$

Therefore, the two lines are perpendicular to each other.

Question 12: Find the shortest distance between the lines

$\overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda (\widehat{i}-\widehat{j}+\widehat{k})$ and $\overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu (2\widehat{i}+\widehat{j}+2\widehat{k})$

Answer:

So given equation of lines;

$\overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda(\widehat{i}-\widehat{j}+\widehat{k})$and$\overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu (2\widehat{i}+\widehat{j}+2\widehat{k})$ in the vector form.

Now, we can find the shortest distance between the lines $\vec{r} = \vec{a_{1}}+\lambda\vec{b_{1}}$ and $\vec{r} = \vec{a_{2}}+\mu \vec{b_{2}}$ , is given by the formula,

$d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |$

Now, comparing the values from the equation, we obtain

$\vec{a_{1}}=\widehat{i}+2\widehat{j}+\widehat{k}$$\vec{b_{1}}=\widehat{i}-\widehat{j}+\widehat{k}$

$\vec{a_{2}}=2\widehat{i}-\widehat{j}-\widehat{k}$$\vec{b_{2}} = 2\widehat{i}+\widehat{j}+2\widehat{k}$

$\vec{a_{2}} -\vec{a_{1}} =\left ( 2\widehat{i}-\widehat{j}-\widehat{k} \right ) - \left ( \widehat{i}+2\widehat{j}+\widehat{k} \right ) = \widehat{i}-3\widehat{j}-2\widehat{k}$

Then calculating

$\vec{b_{1}}\times \vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 & -1 &1 \\ 2& 1 &2 \end{vmatrix}$

$\vec{b_{1}}\times \vec{b_{2}} = (-2-1)\widehat{i} - (2-2) \widehat{j} +(1+2) \widehat{k} = -3\widehat{i}+3\widehat{k}$

$\Rightarrow \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{(-3)^2+(3)^2} = \sqrt{9+9} =\sqrt{18} =3\sqrt2$

So, substituting the values now in the formula above we get;

$d =\left | \frac{\left ( -3\widehat{i}+3\widehat{k} \right ).(\widehat{i}-3\widehat{j}-2\widehat{k})}{3\sqrt2} \right |$

$\Rightarrow d = \left | \frac{-3.1+3(-2)}{3\sqrt2} \right |$

$d = \left | \frac{-9}{3\sqrt2} \right | = \frac{3}{\sqrt2} = \frac{3\sqrt2}{2}$

Therefore, the shortest distance between the two lines is $\frac{3\sqrt2}{2}$ units.

Question 13: Find the shortest distance between the lines

$\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

Answer:

We have given two lines:

$\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$

Calculating the shortest distance between the two lines,

$\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ and $\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$

by the formula

$d = \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1} &b_{1} &c_{1} \\ a_{2}&b_{2} &c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^2+(c_{1}a_{2}-c_{2}a_{1})^2+(a_{1}b_{2}-a_{2}b_{1})^2}}$

Now, comparing the given equations, we obtain

$x_{1} = -1,\ y_{1} =-1,\ z_{1} =-1$

$a_{1} = 7,\ b_{1} =-6,\ c_{1} =1$

$x_{2} = 3,\ y_{2} =5,\ z_{2} =7$

$a_{2} = 1,\ b_{2} =-2,\ c_{2} =1$

Then, calculating the determinant

$\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1} &b_{1} &c_{1} \\ a_{2}&b_{2} &c_{2} \end{vmatrix} = \begin{vmatrix} 4 &6 &8 \\ 7& -6& 1\\ 1& -2& 1 \end{vmatrix}$

$\Rightarrow$$ 4(-6+2)-6(7-1)+8(-14+6)$

$\Rightarrow$$ -16-36-64$

$\Rightarrow$$-116$

Now, calculating the denominator,

$\sqrt{(b_{1}c_{2}-b_{2}c_{1})^2+(c_{1}a_{2}-c_{2}a_{1})^2+(a_{1}b_{2}-a_{2}b_{1})^2}$

$\Rightarrow$$ \sqrt{(-6+2)^2+(1+7)^2+(-14+6)^2}$ $= \sqrt{16+36+64}$

$ \sqrt{116} = 2\sqrt{29}$

So, we will substitute all the values in the formula above to obtain,

$d = \frac{-116}{2\sqrt{29}} = \frac{-58}{\sqrt{29}} = \frac{-2\times29}{\sqrt{29}} = -2\sqrt{29}$

Since distance is always non-negative, the distance between the given lines is

$2\sqrt{29}$ units.

Question 14: Find the shortest distance between the lines whose vector equations are $\overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k})$ and

$\overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k})$

Answer:

Given two equations of line

$\overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+\lambda (\widehat{i}-3\widehat{j}+2\widehat{k})$$\overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k})$ in the vector form.

So, we will apply the distance formula for knowing the distance between two lines $\vec{r} =\vec{a_{1}}+\lambda{b_{1}}$ and $\vec{r} =\vec{a_{2}}+\lambda{b_{2}}$

$d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |$

After comparing the given equations, we obtain

$\vec{a_{1}}=\widehat{i}+2\widehat{j}+3\widehat{k}$$\vec{b_{1}}= \widehat{i}-3\widehat{j}+2\widehat{k}$

$\vec{a_{2}}=4\widehat{i}+5\widehat{j}+6\widehat{k}$$\vec{b_{2}}= 2\widehat{i}+3\widehat{j}+\widehat{k}$

$\vec{a_{2}}-\vec{a_{1}}=(4\widehat{i}+5\widehat{j}+6\widehat{k})- (\widehat{i}+2\widehat{j}+3\widehat{k})$

$= 3\widehat{i}+3\widehat{j}+3\widehat{k}$

Then, calculating the determinant value numerator.

$\vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1& -3 &2 \\ 2& 3& 1 \end{vmatrix}$

$=(-3-6)\widehat{i}-(1-4)\widehat{j}+(3+6)\widehat{k}= -9\widehat{i}+3\widehat{j}+9\widehat{k}$

That implies, $\left | \vec{b_{1}}\times\vec{b_{2}} \right | = \sqrt{(-9)^2+(3)^2+(9)^2}$

$= \sqrt{81+9+81} = \sqrt{171} =3\sqrt{19}$

$\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(-9\widehat{i}+3\widehat{j}+9\widehat{k})(3\widehat{i}+3\widehat{j}+3\widehat{k})$

$= (-9\times3)+(3\times3)+(9\times3) = 9$

Now, after substituting the value in the above formula, we get,

$d= \left | \frac{9}{3\sqrt{19}} \right | = \frac{3}{\sqrt{19}}$

Therefore, $\frac{3}{\sqrt{19}}$ is the shortest distance between the two given lines.

Question 15: Find the shortest distance between the lines whose vector equations are

$\overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k}$ and $\overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}$

Answer:

Given two equations of the line

$\overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k}$ $\overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}$ in the vector form.

So, we will apply the distance formula for knowing the distance between two lines $\vec{r} =\vec{a_{1}}+\lambda{b_{1}}$ and $\vec{r} =\vec{a_{2}}+\lambda{b_{2}}$

$d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |$

After comparing the given equations, we obtain

$\vec{a_{1}}=\widehat{i}-2\widehat{j}+3\widehat{k}$$\vec{b_{1}}=-\widehat{i}+\widehat{j}-2\widehat{k}$

$\vec{a_{2}}=\widehat{i}-\widehat{j}-\widehat{k}$$\vec{b_{2}}= \widehat{i}+2\widehat{j}-2\widehat{k}$

$\vec{a_{2}}-\vec{a_{1}}=(\widehat{i}-\widehat{j}-\widehat{k})- (\widehat{i}-2\widehat{j}+3\widehat{k}) = \widehat{j}-4\widehat{k}$

Then, calculating the determinant value numerator.

$\vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ -1& 1 &-2 \\ 1& 2& -2 \end{vmatrix}$

$=(-2+4)\widehat{i}-(2+2)\widehat{j}+(-2-1)\widehat{k}= 2\widehat{i}-4\widehat{j}-3\widehat{k}$

That implies,

$\left | \vec{b_{1}}\times\vec{b_{2}} \right | = \sqrt{(2)^2+(-4)^2+(-3)^2}$

$= \sqrt{4+16+9} = \sqrt{29}$

$\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(2\widehat{i}-4\widehat{j}-3\widehat{k})(\widehat{j}-4\widehat{k}) = -4+12 = 8$

Now, after substituting the value in the above formula, we get,

$d= \left | \frac{8}{\sqrt{29}} \right | = \frac{8}{\sqrt{29}}$

Therefore, $\frac{8}{\sqrt{29}}$ units is the shortest distance between the two given lines.

Question 1: Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Answer:

Given direction ratios $a,b,c$ and $b-c,\ c-a,\ a-b$ .

Thus, the angle between the lines A is given by;

$A = \left | \frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}.\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} \right |$

$\Rightarrow \cos A = 0$

$\Rightarrow A = \cos^{-1}(0) = 90^{\circ}$

Thus, the angle between the lines is $90^{\circ}$.

Question 2: Find the equation of a line parallel to the x-axis and passing through the origin.

Answer:

Equation of a line parallel to the x-axis and passing through the origin $(0,0,0)$ is itself the x-axis.

So, let A be a point on the x-axis.

Therefore, the coordinates of A are given by $(a,0,0)$, where $a\epsilon R$.

Now, the direction ratios of OA are $(a-0) =a,0 , 0$

So, the equation of OA is given by,

$\frac{x-0}{a} = \frac{y-0}{0} = \frac{z-0}{0}$

or $\Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a$

Thus, the equation of the line parallel to the x-axis and passing through the origin is

$\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$

Question 3: If the lines $\frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular, find the value of k.

Answer:

Given both lines are perpendicular so we have the relation; $a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0$

For the two lines whose direction ratios are known,

$a_{1},b_{1},c_{1}\ and\ a_{2},b_{2},c_{2}$

We have the direction ratios of the lines, $\frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are $-3,2k,2$ and $3k,1,-5$ respectively.

Therefore, applying the formula,

$-3(3k)+2k(1)+2(-5) = 0$

$\Rightarrow -9k +2k -10 = 0$

$\Rightarrow7k=-10$ or $k= \frac{-10}{7}$

$\therefore$ For, $k= \frac{-10}{7}$ the lines are perpendicular.

Question 4: Find,the shortest distance between lines $\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k})$ and $\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k})$ .

Answer:

Given lines are;

$\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k})$ and

$\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k})$

So, we can find the shortest distance between two lines $\vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}}$ and $\vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}}$ by the formula,

$d = \left | \frac{(\vec{b_{1}}\times\vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})}{\left | \vec{b_{1}}\times\vec{b_{2}} \right |} \right |$ ...........................(1)

Now, we have from the comparisons of the given equations of lines.

$\vec{a_{1}}=6\widehat{i}+2\widehat{j}+2\widehat{k}$$\vec{b_{1}}= \widehat{i}-2\widehat{j}+2\widehat{k}$

$\vec{a_{2}}=-4\widehat{i}-\widehat{k}$$\vec{b_{2}}= 3\widehat{i}-2\widehat{j}-2\widehat{k}$

So,$\vec{a_{2}}-\vec{a_{1}}=(-4\widehat{i}-\widehat{k})-(6\widehat{i}+2\widehat{j}+2\widehat{k}) = -10\widehat{i}-2\widehat{j}-3\widehat{k}$

and $\Rightarrow \vec{b_{1}}\times \vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 &-2 &2 \\ 3& -2 &-2 \end{vmatrix} = (4+4)\widehat{i}-(-2-6)\widehat{j}+(-2+6)\widehat{k}$

$=8\widehat{i}+8\widehat{j}+4\widehat{k}$

$\therefore \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{8^2+8^2+4^2} =12$

$(\vec{b_{1}}\times\vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})= (8\widehat{i}+8\widehat{j}+4\widehat{k}).(-10\widehat{i}-2\widehat{j}-3\widehat{k})= -80-16-12 =-108$ Now, substituting all values in equation (3) we get,

$d = | \frac{-108}{12}| = 9$

Hence the shortest distance between the two given lines is 9 units.

Question 5: Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

$\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$and$\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$

Answer:

Given

Two straight lines in 3D whose direction cosines are (3,-16,7) and (3,8,-5)

Now, the two vectors which are parallel to the two lines are

$\vec a= 3\hat i-16\hat j+7\hat k$ and

$\vec b= 3\hat i+8\hat j-5\hat k$

As we know, a vector perpendicular to both vectors $\vec a$ and $\vec b$ is $\vec a\times\vec b$ , so

$\vec a\times\vec b=\begin{vmatrix} \hat i &\hat j &\hat k \\ 3& -16 &7 \\ 3&8 &-5 \end{vmatrix}=\hat i(80-56)-\hat j(-15-21)+\hat k(24+48)$

$\vec a\times\vec b=24\hat i+36\hat j+72\hat k$

A vector parallel to this vector is

$\vec d=2\hat i+3\hat j+6\hat k$

Now, as we know, the vector equation of the line which passes through point p and is parallel to vector d is

$L=\vec p+\lambda \vec d$

Here in our question, give point p = (1,2,-4), which means the position vector of this point is

$\vec p = \hat i +2\hat j-4\hat k$

So, the required line is

$L=\vec p+\lambda \vec d$

$\Rightarrow$$L=\hat i+2\hat j-4\hat k +\lambda (2\hat i+3\hat j+6\hat k)$

$\Rightarrow$$L=(2\lambda +1)\hat i+(2+3\lambda)\hat j+(6\lambda-4)\hat k$