ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.2

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ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.2

Very short answer type questions (1 to 7) :

Question 1.
(i) Find the vector equation of the line passing through the point (3, 4, 5) and parallel to the vector \(2 \hat{i}+2 \hat{j}-3 \hat{k}\). (NCERT)
(ii) Find the vector equation of the line which is parallel to the vector \(3 \hat{i}-2 \hat{j}+6 \hat{k}\) and passes through the point (1, – 2, 3). (NCERT Exemplar)
Solution:
(i) Clearly the vector eqn. of line through point (3, 4, 5) whose position vector \(\vec{a}=3 \hat{i}+4 \hat{j}+5 \hat{k}\)
and parallel to the vector \(\vec{b}=2 \hat{i}+2 \hat{j}-3 \hat{k}\) be given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\) where λ be the parameter
⇒ \(\vec{r}=(3 \hat{i}+4 \hat{j}+5 \hat{k})+\lambda(2 \hat{i}+2 \hat{j}-3 \hat{k})\)
where \(\vec{r}\) be the P.V. of point on required line.

(ii) Let \(\vec{r}\) be the position vector of the point P (x, y, z) on the given line.
Then vector eqn. of line through the point whose P.V \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\)
and parallel to vector \(\vec{b}=3 \hat{i}-2 \hat{j}+6 \hat{k}\) is given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\)
i.e. \(\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k})\)

Question 1 (old).
(i) Find the vector equation of the line passing through the point (1, 2, 3) and parallel to the vector \(3 \hat{i}+2 \hat{j}-2 \hat{k}\). (NCERT)
Solution:
Let \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)
and \(\vec{b}=3 \hat{i}+2 \hat{j}-2 \hat{k}\)
∴ required vector eqn. of line is given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\)
i.e. \(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}-2 \hat{k})\).

Question 2.
Find the cartesian equation of the line which passes through the point with position vector \(2 \hat{i}-\hat{j}+4 \hat{k}\) and is in the direction of the vector \(\hat{i}+2 \hat{j}-\hat{k}\).
Solution:
Here \(\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}\)
and \(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\)
∴ vector eqn. of line be given by \(\vec{r}=\vec{a}+\lambda \vec{b}\)
∴ \(\vec{r}=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})\)
⇒ \(x \hat{i}+y \hat{j}+z \hat{k}=(2+\lambda) \hat{i}+(-1+2 \lambda) \hat{j}+(4-\lambda) \hat{k}\)
On comparing ;
x = 2 + λ ;
y = – 1 + 2λ ;
z = 4 – λ
⇒ \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}\) (= λ)
be the required eqn. of line in cartesian form.

Question 3.
(i) Find the vector equation of the line \(\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}\). (NCERT)
(ii) Write the vector equation of hte line \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{6-z}{2}\).
Solution:
(i) The given line \(\frac{x-(-3)}{2}=\frac{y-5}{4}=\frac{z-(-6)}{2}\)
given line passes through the point (- 3, 5, – 6)
whose P.V. be \(\vec{a}=-3 \hat{i}+5 \hat{j}-6 \hat{k}\)
and parallel to vector \(\vec{b}=2 \hat{i}+4 \hat{j}+2 \hat{k}\)
i.e. \(\vec{a}=-3 \hat{i}+5 \hat{j}-6 \hat{k}\)
and \(\vec{b}=2 \hat{i}+4 \hat{j}+2 \hat{k}\)
∴ Vector eqn. of line be, \(\vec{r}=\vec{a}+\lambda \vec{b}\).

(ii) Cartesian eqn. of given line be
\(\frac{x-5}{3}=\frac{y-(-4)}{7}=\frac{z-6}{2}\) = t (say)
⇒ x = 3t + 5 ;
y = 7t – 4 ;
z = 2t + 6
Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) be the P.V of any point on the line
Then \(\vec{r}=(3 t+5) \hat{i}+(7 t-4) \hat{j}+(2 t+6) \hat{k}\)
⇒ \(\vec{r}=5 \hat{i}-4 \hat{j}+6 \hat{k}+t(3 \hat{i}+7 \hat{j}+2 \hat{k})\) be the required eqn. of line in vector form.

Question 4.
(i) Write the cartesian equation of the following line given in vector form :
\(\vec{r}=2 \hat{i}+\hat{j}-4 \hat{k}+\lambda(\hat{i}-\hat{j}-\hat{k})\).
(ii) Find the cartesian equation of the line which passes through the point (- 2, 4, – 5) and parallel to the line \(\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}\).
Solution:
(i) Given vector eqn. of line be
\(\vec{r}=2 \hat{i}+\hat{j}-4 \hat{k}+\lambda(\hat{i}-\hat{j}-\hat{k})\)
Now \(\vec{r}\) be the position vector of P (x, y, z), we have
\(x \hat{i}+y \hat{j}+z \hat{k}=2 \hat{i}+\hat{j}-4 \hat{k}+\lambda(\hat{i}-\hat{j}-\hat{k})\)
⇒ \(x \hat{i}+y \hat{j}+z \hat{k}=(2+\lambda) \hat{i}+(1-\lambda) \hat{j}+(-4-\lambda) \hat{k}\)
Comparing the coefficients of \(\hat{i}, \hat{j}, \hat{k}\) ; we have
x = 2 + λ
⇒ x – 2 = λ
y = 1 – λ
⇒ 1 – y = λ
z = – 4 – λ
⇒ – (z + 4) = λ
On eliminating the parameter λ ; we have
⇒ \(\frac{x-2}{1}=\frac{y-1}{-1}=\frac{z+4}{-1}\), which is required eqn. of line in cartesian form.

(ii) eqn. of given line,
\(\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}\)
i.e. \(\frac{x+3}{3}=\frac{y-4}{-5}=\frac{z+8}{6}\) ……………….(1)
∴ D’ ratios of given line (1) be < 3, – 5, 6 >
Thus eqn. of line passing through the point (- 2, 4, – 5)
and having direction ratios < 3, – 5, 6 > is given by \(\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z+5}{6}\).

Question 5.
(i) Find the vector equation of the line passing through the points (- 1, 0, 2) and (3, 4, 6). (NCERT)
(ii) Find the vector and the cartesian equations of the line that passes through origin and the point (5, – 2, 3). (NCERT)
Solution:
(i) Let \(\vec{a} \text { and } \vec{b}\) are the position vectors of the given points
∴ \(\vec{a}=-\hat{i}+2 \hat{k}\)
and \(\vec{b}=3 \hat{i}+4 \hat{j}+6 \hat{k}\)
i.e. \(\vec{b}-\vec{a}=4 \hat{i}+4 \hat{j}+4 \hat{k}\)
∴ Vector eqn. of line be \(\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\)
i.e. \(\vec{r}=(-\hat{i}+2 \hat{k})+\lambda(4 \hat{i}+4 \hat{j}+4 \hat{k})\)
where λ be any scalar.

(ii) Let \(\vec{a} \text { and } \vec{b}\) be the position vectors of the given points
i.e. \(\vec{a}=0 \hat{i}+0 \hat{j}+0 \hat{k}\)
and \(\vec{b}=5 \hat{i}-2 \hat{j}+3 \hat{k}\)
∴ \(\vec{b}-\vec{a}=5 \hat{i}-2 \hat{j}+3 \hat{k}\)
Let \(\vec{r}\) be the position vector of any point on the line
∴ Vector eqn. of line be \(\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\)
i.e. \(\vec{r}=\overrightarrow{0}+\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})\)
⇒ \(x \hat{i}+y \hat{j}+z \hat{k}=(0 \hat{i}+0 \hat{j}+0 \hat{k})+\lambda(5 \hat{i}-2 \hat{j}+3 \hat{k})\)
i.e. x = 5λ ;
y = – 2λ ;
z = 3λ
⇒ \(\frac{x-0}{5}=\frac{y-0}{-2}=\frac{z-0}{3}\) (= λ) be the required eqn. of line in cartesian form.

Question 6.
(i) 1f the equation of a line AB is \(\frac{x-3}{1}=\frac{y+2}{-2}=\frac{z-5}{4}\), find the direction ratios of a line parallel to AB.
(ii) The cartesian equations of a line AB are \(\frac{2 x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}\). Write the direction ratios of a line parallel to AB.
(iii) Write the direction cosines of a vector parallel to the line \(\frac{4-x}{2}=\frac{y+3}{3}=\frac{z+2}{6}\).
(iv) The cartesian equation of a line AB is \(\frac{2 x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}\). Find the direction cosines of a line parallel to AB.
Solution:
(i) Given eqn. of line AB
\(\frac{x-3}{1}=\frac{y+2}{-2}=\frac{z-5}{4}\)
∴ D’ ratios of the line AB are < 1, – 2, 4 >
∴ D’ ratios of the line parallel to AB are < 1, – 2, 4 >

(ii) eqns. of given line AB are ;
\(\frac{2 x-1}{2}=\frac{4-y}{7}=\frac{z+1}{2}\) i.e. \(\frac{x-1 / 2}{1}=\frac{y-4}{-7}=\frac{z-(-1)}{2}\)
∴ D’ ratios of line AB are < 1, – 7, 2 >
∴ D’ratios of the line || to AB are < 1, – 7, 2 >

(iii) Given line in cartesian form be given by \(\frac{4-x}{2}=\frac{y+3}{3}=\frac{z+2}{6}\)
i.e. \(\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}\) ………………(1)
Direction ratios of line (1) are proportional to < – 2, 6, – 3>
Thus direction cosines of line (1) is given by
< \(\frac{-2}{\sqrt{(-2)^2+6^2+(-3)^2}}, \frac{6}{\sqrt{(-2)^2+6^2+(-3)^2}}, \frac{-3}{\sqrt{(-2)^2+6^2+(-3)^2}}\) >
i.e. < \(-\frac{2}{7}, \frac{6}{7},-\frac{3}{7}\) >
Since line given by eqn. (1) passes through the point (4, 0, 1) and having direction ratios proportional to < – 2, 6, – 3 >.
Thus in vector form, it means that the line passing through the point with P.V. \(\vec{a}=4 \hat{i}+0 \hat{j}+\hat{k}\)
and parallel to vector \(\vec{b}=-2 \hat{i}+6 \hat{j}-3 \hat{k}\)
Thus required direction cosines of line parallel to line (1) are < \(\frac{-2}{7}, \frac{6}{7}, \frac{-3}{7}\) >.

(iv) Thus required direction cosines of line || to line (1) are < \(\frac{-2}{7}, \frac{6}{7}, \frac{-3}{7}\) >
\(\frac{x-1 / 2}{\sqrt{3} / 2}=\frac{y-(-2)}{2}=\frac{z-3}{3}\)
∴ Direction ratios of line are < \(\frac{\sqrt{3}}{2}\), 2, 3 >
Direction cosines of line are < \(\frac{\frac{\sqrt{3}}{2}}{\sqrt{\frac{3}{4}+4+9}}, \frac{2}{\sqrt{\frac{3}{4}+4+9}}, \frac{3}{\sqrt{\frac{3}{4}+4+9}}\) >
i.e. < \(\frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}\) >

Question 7.
(i) The cartesian equations of a line are 3x – 3 = 2y + 1 = 5 – 6z. Write the direction ratios of the line.
(ii) Find the vector equationof the X-axis.
Solution:
(i) Given eqn. of line be 3x – 3 = 2y + 1 = 5 – 6z
⇒ \(\frac{x-1}{\frac{1}{3}}=\frac{y+\frac{1}{2}}{\frac{1}{2}}=\frac{z-\frac{5}{6}}{-\frac{1}{6}}\)
⇒ \(\frac{x-1}{2}=\frac{y+\frac{1}{2}}{3}=\frac{z-\frac{5}{6}}{-1}\)
∴ D’ ratios of the line are < 2, 3, – 1 >.

(ii) Since X-axis is passing through (0. 0, 0) and having direction ratios < 1, 0, 0 >.
∴ vector eqn. of X-axis is given by \(\vec{r}=0 \hat{i}+0 \hat{j}+0 \hat{k}+\lambda(\hat{i})\)
i.e. \(\vec{r}=\lambda \hat{i}\)
where λ be the parameter.

Question 8.
(i) Find the vector and the cartesian equations of the line through the point (5, 2, – 4) and which is parallel to the vector \(3 \hat{i}+2 \hat{j}-8 \hat{k}\). (NCERT)
(ii) A line passes through the point with position vector \(2 \hat{i}-\hat{j}+4 \hat{k}\) and is in the direction of \(\hat{i}+\hat{j}-2 \hat{k}\). Find the equations of the line in vector and in cartesian form. (NCERT)
Solution:
(i) Let \(\vec{a}=5 \hat{i}+2 \hat{j}-4 \hat{k}\)
and \(\vec{b}=3 \hat{i}+2 \hat{j}-8 \hat{k}\)
Let \(\vec{r}\) be the position vector of any point on the required line.
Vector eqn. of line be \(\vec{r}=\vec{a}+\lambda \vec{b}\)
i.e. \(\vec{r}=(5 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(3 \hat{i}+2 \hat{j}-8 \hat{k})\)
⇒ \(x \hat{i}+y \hat{j}+z \hat{k}=(5+3 \lambda) \hat{i}+(2+2 \lambda) \hat{j}+(-4-8 \lambda) \hat{k}\)
On comparing ;
x = 5 + 3λ ;
y = 2 + 2λ ;
z = – 4 – 8λ
⇒ \(\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}\) (= λ) be the required eqn. in cartesian form.

(ii) Let \(\vec{r}\) be the position vector of point P x, y, z) on given line.
Then vector eqn. of line passing through the point whose P.V. \(\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}\)
and parallel to \(\vec{b}=\hat{i}+\hat{j}-2 \hat{k}\) is given by \(\vec{r}=\vec{a}+\lambda \vec{b}\)
⇒ \(\vec{r}=2 \hat{i}-\hat{j}+4 \hat{k}+\lambda(\hat{i}+\hat{j}-2 \hat{k})\) …………………….(1)
Since \(\vec{r}\) be the P.V of point (x, z).
Then \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ eqn..(1) becomes ;
\(x \hat{i}+y \hat{j}+z \hat{k}=2 \hat{i}-\hat{j}+4 \hat{k}+\lambda(\hat{i}+\hat{j}-2 \hat{k})\)
⇒ \(x \hat{i}+y \hat{j}+z \hat{k}=(2+\lambda) \hat{i}+(-1+\lambda) \hat{j}+(4-2 \lambda) \hat{k}\)
On comparing the coefficients of \(\hat{i}, \hat{j} \text { and } \hat{k}\) on both sides ; we have
x = 2 + λ
⇒ x – 2 = λ ;
y = – 1 + λ
⇒ y + 1 = λ;
z = 4 – 2λ
⇒ \(\frac{z-4}{-2}\) = λ
On eliminating λ from these eqn’s ; we have
\(\frac{x-2}{1}=\frac{y+1}{1}=\frac{z-4}{-2}\)
which is the required cartesian eqn. of line.

Question 9.
(i) Find the coordinates of the point where the line \(\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}\) cuts the xy-plane.
(ii) Find the coordiantes of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the zx-plane. (NCERT)
Solution:
Eqn. of given line
\(\frac{x-1}{3}=\frac{y+4}{7}=\frac{z+4}{2}\) = t (say)
anypoint on line (1) be P(3t + 1, 7t – 4, 2t – 4)
and it is given that line (1) meets xy plane.
i.e. z-coordinate of point P is 0.
∴ 2t – 4 = 0
⇒ t = 2
Thus required coordinates of point P are (3 × 2 + 1, 7 × 2 – 4, 2 × 2 – 4) i.e. (7, 10, 0).

(ii) Direction ratios of the line AB is given by < 3 – 5, 4 – 1, 1 – 6 >
i.e. < – 2, 3, – 5>
Thus, eqn. of line through the points A (5, 1, 6) and B (3, 4, 1) is given by
\(\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}\) = t (say)
So any point on given line be (- 2t + 5, 3t + 1, – 5t + 6)
Since the line crosses the zx-plane
∴ y-coordinate of any point is 0.
∴ 3t + 1 = 0
⇒ t = – \(\frac{1}{3}\)
Thus, required point be, (\(\frac{2}{3}\) + 5, 0, \(\frac{5}{3}\) + 6) i.e. (\(\frac{17}{3}\), 0, \(\frac{23}{3}\))

Question 11.
The carteian equations of a line are x = ay + b, z = cy + d. Find direction numbers of the line, also find its equation in vector form.
Solution:
Given cartesian eqn. of line be,
x = ay + b,
z = cy + d
⇒ \(\frac{x-b}{a}\) = y ;
\(\frac{z-d}{c}\) = y
⇒ \(\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c}\) = t (say)
Here direction ratios of given line are proportional to < a, 1, c >.
∴ x = at + b ;
y = t ;
z = ct + d
Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) be the RV. of any point on given line
Then \(\vec{r}=(a t+b) \hat{i}+t \hat{j}+(c t+d) \hat{k}\)
⇒ \(\vec{r}=(b \hat{i}+0 \hat{j}+d \hat{k})+t(a \hat{i}+\hat{j}+c \hat{k})\)
where t be the parameter be the required vector eqn. of line.

Question 12.
(i) Find the vector equation of the line passing through the point A (2, – 1, 1) and parallel to the line joining the points B (- 1, 4,1) and C (1, 2, 2). Also find the cartesian equation of the line.
(ii) Find the vector equation of a line passing through the point with position vector \(\hat{i}-2 \hat{j}-3 \hat{k}\) and parallel to the line joining the points with position vectors \(\hat{i}-\hat{j}+4 \hat{k}\) and \(2 \hat{i}+\hat{j}+2 \hat{k}\). Also find the cartesian equations of the line.
Solution:
(i) D’ ratios of line BC are < 1 + 1, 2 – 4, 2 – 1 > i.e. < 2, – 2, 1 >
Let \(\vec{r}\) be the position vector of point P (x, y, z) on given line.
Then eqn. of line through the point whose position vector \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\) and || to line BC is given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\)
⇒ \(\vec{r}=2 \hat{i}-\hat{j}+\hat{k}+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})\)
Since \(\vec{r}\) be the P.V of point P (x, y, z).
Then \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
∴ from (1) ; we have
\(x \hat{i}+y \hat{j}+z \hat{k}=(2+2 \lambda) \hat{i}+(-1-2 \lambda) \hat{j}+(1+\lambda) \hat{k}\)
Comparing the coefficients of \(\hat{i}, \hat{j}, \hat{k}\) on both sides, we have
x = 2 + 2λ
⇒ \(\frac{x-2}{2}\) = λ ;
y = – 1 – 2λ
⇒ \(\frac{y+1}{-2}\) = λ ;
z = 1 + λ
⇒ z – 1 = λ
eliminating λ from these eqn’s, we have \(\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-1}{1}\)
which gives the required cartesian eqn. of line.

(ii) Let A, B, C be the given points whose position vectors are
\(\hat{i}-2 \hat{j}-3 \hat{k}\) ;
\(\hat{i}-\hat{j}+4 \hat{k}\) ;
\(2 \hat{i}+\hat{j}+2 \hat{k}\)
we want to find the eqn. of line passing through A and parallel to \(\overrightarrow{\mathrm{BC}}\).
Now \(\overrightarrow{\mathrm{BC}}\) = P.V.of C – P.V.of B
= \((2 \hat{i}+\hat{j}+2 \hat{k})-(\hat{i}-\hat{j}+4 \hat{k})\)
= \(\hat{i}+2 \hat{j}-2 \hat{k}\)
also vector eqn. of Line passes through point A with PV, \(\vec{a}\) and || to \(\vec{b}\) is given by
\(\vec{r}=\vec{a}+\lambda \vec{b}\)
Here \(\vec{a}=\hat{i}-2 \hat{j}-3 \hat{k}\)
and \(\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
i.e. \(\vec{r}=\hat{i}-2 \hat{j}-3 \hat{k}+\lambda(\hat{i}+2 \hat{j}-2 \hat{k})\) be the required vector equation.
Putting \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) in this line, we get
\(x \hat{i}+y \hat{j}+z \hat{k}=(1+\lambda) \hat{i}+(-2+2 \lambda) \hat{j}+(-3-2) \hat{k}\)
equating the coefficients \(\mathrm{f} \hat{i}, \hat{j}, \hat{k}\) on both sides, we get
∴ x = 1 + λ ;
y = – 2 + 2λ ;
z = – 3 – 2λ
⇒ \(\frac{x-1}{1}\) = λ ;
\(\frac{y+2}{2}\) = λ ;
\(\frac{z+3}{-2}\) = λ
On eliminating λ we get
\(\frac{x-1}{1}=\frac{y+2}{2}=\frac{z+3}{-2}\) be the required cartesian equation of line.

Question 12 (old).
The cartesian equations of a line are 6x – 2 = 3y + 1 = 2z – 2. Find direction ratios of the line and write down the vector equation of the line through (2, – 1, – 1) which is parallel to the given line.
Solution:
The given line in cartesian form be 6x – 2 = 3y + 1 = 2z – 2.
i.e. \(\frac{x-1 / 3}{1 / 6}=\frac{y+1 / 3}{1 / 3}=\frac{z-1}{1 / 2}\)
i.e. \(\frac{x-1 / 3}{1}=\frac{y+1 / 3}{2}=\frac{z-1}{3}\)

(a) ∴ D’ ratios of line are < 1, 2, 3 >

(b) Here \(\vec{a}=(2 \hat{i}-\hat{j}-\hat{k})\)
Since we want to find the eqn. of line through the point whose P.V. be \(\vec{a}\) and || to given line.
∴ given line be || to vector \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\)
∴ vector eqn. of line be \(\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})\).

Question 13.
Find the vector equation of the line joining the points whose position vectors are \(2 \hat{i}-\hat{j}+\hat{k}\) and \(\hat{i}+2 \hat{j}-3 \hat{k}\). Also find its cartesian equations.
Solution:
Let \(\vec{a}, \vec{b}\) are the position vectors of the points. Then
∴ \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}\)
∴ \(\vec{b}-\vec{a}=-\hat{i}+3 \hat{j}-4 \hat{k}\)
Let \(\vec{r}\) be the position vector of any point on the line
∴ vector eqn. of line is \(\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\)
i.e. \(\vec{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(-\hat{i}+3 \hat{j}-4 \hat{k})\)
Since \(\vec{r}\) be P.V. of any point P (x, y, z) on line.
⇒ \(x \hat{i}+y \hat{j}+z \hat{k}=(2-\lambda) \hat{i}+(-1+3 \lambda) \hat{j}+(1-4 \lambda) \hat{k}\)
On comparing ;
x = 2 – λ ;
y = – 1 + 3λ ;
z = 1 – 4λ
⇒ \(\frac{x-2}{-1}=\frac{y+1}{3}=\frac{z-1}{-4}\) (= λ) be the required eqn. in cartesian form.

Question 14.
Find the vector equation of the line passing through the point A (1, 2, – 1) and parallel to the line 5x – 25 = 14 – 7y = 35z.
Solution:
eqn. of given line 5x – 25 = 14 – 7y = 35z
⇒ 5 (x – 5) = – 7 (y – 2) = 35z
⇒ \(\frac{x-5}{7}=\frac{(y-2)}{-5}=\frac{z}{1}\) ……………..(1)
Direction ratios of line (1) are < 7, – 5, 1 >
Since required line is parallel to line (1)
∴ direction ratios of line parallel to line (1) are < 7, – 5, 1 >.
Thus, required eqn. of line passing through the point A (1, 2, – 1) whose position vector \(\vec{a}=\hat{i}+2 \hat{j}-\hat{k}\)
and || to vector \(\vec{b}=7 \hat{i}-5 \hat{j}+\hat{k}\) be given by \(\vec{r}=\vec{a}+\lambda \vec{b}\)
⇒ \(\vec{r}=(\hat{i}+2 \hat{j}-\hat{k})+\lambda(7 \hat{i}-5 \hat{j}+\hat{k})\)
where λ be the parameter.

Question 14 (old).
(i) Find the coordinates of the point where the line through A (3, 4, 1) and B (5, 1, 6) crosses the y-pIane.
(iii) If the x-coordinate of a point on the line joining the points P (2, 2, 1) and Q (5, 1, – 2) is 4, find its z-coordinate.
Solution:
(i) Direction ratios of the line AB be < 5 – 3, 1 – 4, 6 – 1 >
i.e. < 2, – 3, 5 >
∴ eqn. of line through the points A (3, 4, 1) and B (5, 1,6) is given by
\(\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}\) = t (say)
Any point on given line be (2t + 3, – 3t + 4, 5t + 1)
Since the line AB crosses the xy-plane
∴ Z-coordinate of any point is 0.
∴ 5t + 1 = 0
⇒ t = – \(\frac{1}{5}\)
∴ required coordinates of any point on line be \(\left(-\frac{2}{5}+3, \frac{3}{5}+4,0\right)\)
i.e. \(\left(\frac{13}{5}, \frac{23}{5}, 0\right)\).

(iii) The equation of line joining P (2, 2, 1) and Q (5, 1, – 2) is given by \(\frac{x-2}{5-2}=\frac{y-2}{1-2}=\frac{z-1}{-2-1}\)
i.e. \(\frac{x-2}{3}=\frac{y-2}{-1}=\frac{z-1}{-3}\) = t (say)
∴ Any point on line PQ be (31 + 2, – t + 2, – 3, + 1)
Since it is given that x-coordinate of point on line PQ be 4.
∴ 3t + 2 = 4
⇒ t = \(\frac{2}{3}\)
∴ required z-coordinate = – 3t + 1
= – 3 (\(\frac{2}{3}\)) + 1 = – 1.

Question 15.
The points A (4, 5, 10), B (2, 3, 4) and C (1, 2, – 1) are three vertices of a is parallelogram ABCD. Find the vector equations of the sides AB and BC. Also find the coordinates of D.
Solution:
Let \(\vec{a}, \vec{b}, \vec{c} \text { and } \vec{d}\) are the position vectors of given points
A (4, 5, 10), B (2, 3, 4), C (1, 2, – 1) and D (α, β, γ)
∴ \(\vec{a}=4 \hat{i}+5 \hat{j}+10 \hat{k}\) ;
\(\vec{b}=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
\(\vec{c}=\hat{i}+2 \hat{j}-\hat{k}\)
and \(\vec{d}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\)

Question 16.
(i) Find the equation of the line passing through the points A (- 2, 4, 7), B (3, – 6, – 8). Hence, show that the points A, B and C(1, —2, —2) are collinear.
(ii) 1f the points (- 1, 3, 2), (- 4, 2, – 2) and (5, 5, λ) are collinear, then find the value of λ.
Solution:
(i) Direction ratios of given line AB is < 3 + 2, – 6 – 4, – 8 – 7 > i.e.< 5, – 10, – 15 > i.e.< 1, – 2, – 3 >
∴ required eqn. of line passing through the points A (- 2, 4, 7) and B (3, – 6, – 8) is given by
\(\frac{x+2}{1}=\frac{y-4}{-2}=\frac{z-7}{-3}\) ……………..(1)
The point C (1, – 2, – 2) lies on eqn. (1)
if \(\frac{1+2}{1}=\frac{-2-4}{-2}=\frac{-2-7}{-3}\) if 3 = 3 = 3, which is true
∴ point C (1, – 2, – 2) lies on line AB.
∴ the points A, B and C lies on same line.
Thus, points A, B, C are collinear.

(ii) Direction ratios of line through the points (- 1, 3, 2) and (4, 2, – 2) be < – 4 + 1, 2 – 3, – 2 – 2 >
i.e. < – 3, – 1, – 4 > i.e. < 3, 1, 4 >
∴ eqn. of line through the points (- 1, 3, 2) and (- 4, 2, – 2) is given by
\(\frac{x+1}{3}=\frac{y-3}{1}=\frac{z-2}{4}\) …………….(1)
Now the point (5, 5, λ) lies on eqn. (1),
since (- 1, 3, 2), (- 4, 2, – 2) and (5, 5, λ) are collinear
so all these points lies on same line.
∴ from (1) ;
\(\frac{5+1}{3}=\frac{5-3}{1}=\frac{\lambda-2}{4}\)
2 = \(\frac{\lambda-2}{4}\)
⇒ λ – 2 = 8
⇒ λ = 10.

Question 17.
(i)Find the coordinates of the points on the line \(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6}\) which are at a distance of 3√2 units from the point (1, – 2, 3).
(ii) Find a point on the line \(\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}\) at a distance of 32 units from the point (1, 2, 3).
Solution:
(i) eqn. of given line be, \(\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{6}\) = t (say)
So any point on given line be P (2t + 1, 3t – 2, 6t + 3)
Now it is given that point Pis at a distance of 3 units from given point A (1,- 2, 3)
∴ |AP| = 3

(ii) ∴ The points A, B, C are lies on same line.Hence the points A, B, C are collinear.
Eqn. of given line be
\(\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}\) = t (say) ……………..(1)
Any point on line (1) is given by P(3 – 2, 2t – 1, 2t + 3)
Also |AP| = 3√2 where A (1, 2, 3) be the given point
∴ \(\sqrt{(3 t-2-1)^2+(2 t-1-2)^2+(2 t+3-3)^2}\) = 3√2
⇒ \(\sqrt{(3 t-3)^2+(2 t-3)^2+4 t^2}\) = 3√2
On squaring; we have
9 (t² – 2t + 1) + (4t² + 9 – 12t) + 4t² = 18
⇒ 17t² – 30t = 0
⇒ t = 0, \(\frac{30}{17}\)
When t = 0, point P becomes (- 2, – 1, 3)
When t = 30/17, point P becomes \(\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)\).

Question 18.
Find the points on the line through the points A (3, 4, – 1) and B (5, 1, 5) at a distance of 7 units from the mid-point of line segment AB.
Solution:
Directionratios of line AB be < 5 – 3, 1 – 4, 5 + 1 > i.e. < 2, – 3, 6 >
∴ eqn. of line passing through the points A (3, 4, – 1) and B (5, 1, 5) is given by
\(\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z+1}{6}\) ………………(1)
So Any point on line (1) be P (2t + 3, – 3t + 4, 6t – 1)
Now mid point of line AB be \(\left(\frac{3+5}{2}, \frac{4+1}{2}, \frac{-1+5}{2}\right)\) i.e. M (4, \(\frac{5}{2}\), 2)
Also it is given that |PM| = 7 units
\(\sqrt{(2 t+3-4)^2+\left(-3 t+4-\frac{5}{2}\right)^2+(6 t-1-2)^2}\) = 7
⇒ \(\sqrt{(2 t-1)^2+\left(-3 t+\frac{3}{2}\right)^2+(6 t-3)^2}\) = 7
On squaring; we have
(2t – 1)² + \(\left(-3 t+\frac{3}{2}\right)^2\) + (6t – 3)² = 49
⇒ \(\frac{1}{4}\) [4 (2t – 1)² + (- 6t + 3)² + 4 (6t – 3)²] = 49
⇒ [4 (2t – 1)² + 5 (6t – 3)² + 4 (6t – 3)²] = 196
⇒ 4 (4t² – 4t + 1) + 5 (36t² – 36t + 9) = 196
⇒ 196t² – 196t + 49 = 196
⇒ 4t² – 4t + 1 = 4
⇒ 4t² – 4t – 3 = 0
⇒ t = \(\frac{4 \pm \sqrt{16+48}}{8}\)
= \(\frac{4 \pm 8}{8}\)
t = \(\frac{3}{2}\), – \(\frac{1}{2}\)
When t = \(\frac{3}{2}\)
∴ required point on given line be (3 + 3, – \(\frac{9}{2}\) + 4, 9 – 1) i.e. (6, – \(\frac{1}{2}\), 8)
When t = – \(\frac{1}{2}\)
∴ required point on given line be (- 1 + 3, \(\frac{3}{2}\) + 4, – 3 – 1) i.e. (2, \(\frac{11}{2}\), – 4).

Question 19.
(i) Show that the line \(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\) and \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\) intersect. Also find the coordinates of their point of intersection. (ISC 2017)
(ii) Show that the lines \(\vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})\) ; \(\vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})\) are intersecting. Hence, find their point of intersection.
Solution:
(i) The equations of given lines are
\(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\) ……………….(1)
and \(\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\)
Any point on line (1) be (t + 4,—4t—3, 7:— 1)
it will lies on eqn. (2)
iff \(\frac{t+4-1}{2}=\frac{-4 t-3+1}{-3}=\frac{7 t-1+10}{8}\) are consistent
iff \(\frac{t+3}{2}=\frac{-4 t-2}{-3}\) and \(\frac{-4 t-2}{-3}=\frac{7 t+9}{8}\) are consistent
iff – 3t – 9 = – 8t – 4
and – 32t – 16 = – 21t – 27 are consistent
i.e. iff 5t = 5 and – 11t = – 11 are consistent
i.e. iff t = 1 and t = 1, which is true
Thus both lines intersects and required point of intersection be (1 + 4, – 4 – 3, 7 – 1) i.e. (5, – 7, 6).

(ii) The position vectors of arbitrary points on given lines are;
\(3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})=(3+\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(-4+2 \lambda) \hat{k}\)
and \(5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})=(5+3 \mu) \hat{i}+\hat{j}(-2+2 \mu)+6 \mu \hat{k}\)
If lines are intersects so they have a common point.
So for some values of λ and μ, we must have
3 + λ = 5 + 3μ
⇒ λ – 3μ = 2

2 + 2λ = – 2 + 2μ
⇒ λ – μ = – 2

– 4 + 2λ = 6μ
⇒ λ – 3μ = 2

On solving (1) and (2) ; we have μ = – 2 and λ = – 4 and these values also satisfies eqn. (3).
Clearly given lines intersects.
Putting μ = – 2 in line (2)
we have \(\vec{r}=5 \hat{i}-2 \hat{j}-2(3 \hat{i}+2 \hat{j}+6 \hat{k})\)
⇒ \(\vec{r}=-\hat{i}-6 \hat{j}-12 \hat{k}\) gives the position vector of point of intersection of given lines.
Hence the coordinates of point of intersection be (- 1, – 6, – 12).

Question 20.
(i) Prove that the line through A (0, – 1, – 1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (- 4, 4, 4). Also find their point of intersection. (NCERT Exemplar)
(ii) Show that the lines \(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}\) and \(\frac{x-2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}\) do not intersect.
Solution:
(i) eqn. of line AB is given by \(\frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}\)
i.e. \(\frac{x}{4}=\frac{y+1}{6}=\frac{z+1}{2}\)
i.e. \(\frac{x}{2}=\frac{y+1}{3}=\frac{z+1}{1}\) ……………………(1)

and eqn. of line CD be \(\frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4}\)
i.e. \(\frac{x-3}{-7}=\frac{y-9}{-5}=\frac{z-4}{0}\)
i.e. \(\frac{x-3}{7}=\frac{y-9}{5}=\frac{z-4}{0}\) …………………….(2)
any point on line (1) be given by
\(\frac{x}{2}=\frac{y+1}{3}=\frac{z+1}{1}\) = t (say)
i.e., x = 2t ;
y = 3t – 1 ;
z = t – 1
∴ Coordinates of any point on line (1) be (21, 3t – 1, t – 1) similarly any point online (2) be given by
\(\frac{x-3}{7}=\frac{y-9}{5}=\frac{z-4}{0}\) = s (say)
i.e. x = 7s + 3 ;
y = 5s + 9 ;
z = 4
∴ Coordinates of any point on line (2) be (7s + 3, 5s + 9, 4)
lines (1) and (2) intersects so they have a common point.
So for some values of s and t, we must have
2t = 7s + 3
⇒ 2t – 7s = 3 ……………….(1)

3t 1 = 5s + 9
⇒ 3t – 5s = 10 ………….(2)
t – 1 = 4
⇒ t = 5
On solving (2) and (3);
t = 5 ; s = 1
and these values of s and t satisfies eqn. (1)
Hence both given lines intersects.
putting s = 1 in (7s ± 3, 5s + 9, 4) i.e. (10, 14, 4) be the required point of intersection.

(ii) Comparing the given lines with \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\) and \(\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\)
Here x₁ = 1,
y₁ = – 1,
z₁ = 1 ;

x₂ = 2
y₂ = 1
z₂ = – 1 ;

a₁ = 3,
b₁ = 2,
c₁ = 5

a₂ = 4
b₂ = 3
c₂ = 2

Here \(\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{array}\right|=\left|\begin{array}{ccr}
1 & 2 & -2 \\
3 & 2 & 5 \\
4 & 3 & -2
\end{array}\right|\)
= 1 (- 19) – 2(- 26) – 2(1)
= – 19 + 52 – 2
= 31 ≠ 0
∴ The given lines do not intersects.

Question 21.
A line with direction numbers < 2, 7, – 5 > is drawn to intersect the lines \(\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}\) and \(\frac{x+3}{-3}=\frac{y-3}{2}=\frac{z-6}{4}\).
Find the coordinates of the points of intersection and the length intercepted on it.
Solution:
The eqns on given lines are
\(\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}\) = t (say) ………………….(1)
and \(\frac{x+3}{-3}=\frac{y-3}{2}=\frac{z-6}{4}\) = s (say)

So any point on line (1) be P (3t + 5, – t + 7, t – 2)
andany point on Iine(2) be Q(- 3s – 3, 2s + 3, 4s + 6)
∴ D’ ratios of the line PQ be given by < – 3s – 3 – 3 – 5, 2s + 3 + t – 7, 4s + 6 – t + 2 >
i.e. < – 3s – 3t – 8, 2s + t – 4, 4s – t + 8 >
also direction ratios of line PQ are < 2, 7, – 5>.
∴ \(\frac{-3 s-3 t-8}{2}=\frac{2 s+t-4}{7}=\frac{4 s-t+8}{-5}\)
From (1) and (2) fraction ; we have
– 21s – 21t – 56 = 4s + 2t – 8
⇒ 25s + 231 = – 48 …………………….(3)
From (2) and (3) fraction, we have
– 10s – 5t + 20 = 28s – 7t + 56
⇒ 38s – 2t = – 36
i.e. 19s – t = – 18
On solving eqn. (3) and (4) ; we have
s = – 1 and t = – 1
Thus t = – 1 gives P (2, 8, – 3) and s = – 1
gives Q (0, 1,2)
∴ required length intercepted on given lines = |PQ|
= \(\sqrt{(0-2)^2+(1-8)^2+(2+3)^2}\)
= \(\sqrt{4+49+25}\)
= \(\sqrt{78}\) units.