RD Sharma solution class 12 chapter 28 Straight line in space

Exercise 28.1

Question 1

Find the vector and 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} 

Sol :

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b} 

Here,

\vec{a}=5\hat{i}+2\hat{j}-4\hat{k}

\vec{b}=3\hat{i}+2\hat{j}-8\hat{k}

Vector equation of the required line is given by

\vec{r}=(5\hat{i}+2\hat{j}-4\hat{k})+\lambda(3\hat{i}+2\hat{j}-8\hat{k})     …..1

Here, \lambda is a parameter

Reducing (1) to cartesian form, we get

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(5\hat{i}+2\hat{j}-4\hat{k})+\lambda(3\hat{i}+2\hat{j}-8\hat{k})

\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(5+3\lambda)\hat{i}+(2+2\lambda)\hat{j}+(-4-8\lambda)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\Rightarrow x=5+3\lambda,~~y=2+2\lambda~~z=-4-8\lambda

\Rightarrow \dfrac{x-5}{3}=\lambda,~\dfrac{y-2}{2}=\lambda,~\dfrac{z+4}{-8}=\lambda

Hence, the cartesian form of (1) is

\Rightarrow \dfrac{x-5}{3}=\dfrac{y-2}{2}=\dfrac{z+4}{-8}=\lambda

 

 

Question 2

Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).

Sol :

We know that the vector equation of a line passing through the points with position vectors \vec{a} and parallel to (\vec{b}-\vec{a}) is \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) , where \lambda is some scalar.

Here,

\vec{a}=-1\hat{i}+0\hat{j}+2\hat{k}

\vec{b}=3\hat{i}+4\hat{j}+6\hat{k}

Vector equation of the required line is given by

\Rightarrow \vec{r}=(-1\hat{i}+0\hat{j}+2\hat{k})+\lambda{(3\hat{i}+4\hat{j}+6\hat{k})-(-1\hat{i}+0\hat{j}+2\hat{k})}

\Rightarrow \vec{r}=(-1\hat{i}+0\hat{j}+2\hat{k})+\lambda(4\hat{i}+4\hat{j}+4\hat{k})

Here \lambda is a parameter.

 

 

Question 3

Find the vector equation of a line which is parallel to the vector 2\hat{i}-\hat{j}+3\hat{k} and which passes through the points (5,-2,4) . Also , reduce it to cartesian form.

Sol :

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b} 

Here,

\vec{a}=5\hat{i}-2\hat{j}+4\hat{k}

\vec{b}=2\hat{i}-1\hat{j}+3\hat{k}

Vector equation of the required line is given by

\vec{r}=(5\hat{i}-2\hat{j}+4\hat{k})+\lambda(2\hat{i}-1\hat{j}+3\hat{k})     …..1

Here, \lambda is a parameter

Reducing (1) to cartesian form, we get

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(5\hat{i}-2\hat{j}+4\hat{k})+\lambda(2\hat{i}-1\hat{j}+3\hat{k})

\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(5+2\lambda)\hat{i}+(-2-1\lambda)\hat{j}+(4+3\lambda)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\Rightarrow x=5+2\lambda,~~y=-2-1\lambda~~z=4+3\lambda

\Rightarrow \dfrac{x-5}{2}=\lambda,~\dfrac{y+2}{-1}=\lambda,~\dfrac{z-4}{3}=\lambda

Hence, the cartesian form of (1) is

\Rightarrow \dfrac{x-5}{2}=\dfrac{y+2}{-1}=\dfrac{z-4}{3}=\lambda

 

 

Question 4

A line passes through the point with position vector 2\hat{i}-3\hat{j}+4\hat{k} and is in the direction of 3\hat{i}+4\hat{j}-5\hat{k} .  Find equations of the line in vector and cartesian form.

Sol :

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b} 

Here,

\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}

\vec{b}=3\hat{i}+4\hat{j}-5\hat{k}

Vector equation of the required line is given by

\vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(3\hat{i}+4\hat{j}-5\hat{k})     …..1

Here, \lambda is a parameter

Reducing (1) to cartesian form, we get

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(3\hat{i}+4\hat{j}-5\hat{k})

\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2+3\lambda)\hat{i}+(-3+4\lambda)\hat{j}+(4-5\lambda)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\Rightarrow x=2+3\lambda,~~y=-3+4\lambda~~z=4-5\lambda

\Rightarrow \dfrac{x-2}{3}=\lambda,~\dfrac{y+3}{4}=\lambda,~\dfrac{z-4}{-5}=\lambda

Hence, the cartesian form of (1) is

\Rightarrow \dfrac{x-2}{3}=\dfrac{y+3}{4}=\dfrac{z-4}{-5}=\lambda

 

 

Question 5

ABCD is a parallelogram. The position vectors of the points A,B,C are respectively 4\hat{i}+5\hat{j}-10\hat{k}2\hat{i}-3\hat{j}+4\hat{k} and -\hat{i}+2\hat{j}+\hat{k} . Find the vector equation of the line BD. Also, reduce it to cartesian form.

Sol :

We know that the position vector of the mid-point of \vec{a} and \vec{b} is \dfrac{\vec{a}+\vec{b}}{2}.

Let the position vector of point D be x\hat{i}+y\hat{j}+z\hat{k}

Position vector of mid-point of A and C = Position vector of mid-point of B and D

\Rightarrow \dfrac{(4\hat{i}+5\hat{j}-10\hat{k})+(-\hat{i}+2\hat{j}+\hat{k})}{2}=\dfrac{(2\hat{i}-3\hat{j}+4\hat{k})+(x\hat{i}+y\hat{j}+z\hat{k})}{2}

\Rightarrow \dfrac{3}{2}\hat{i}+\dfrac{7}{2}\hat{j}-\dfrac{9}{2}\hat{k}=\bigg(\dfrac{x+2}{2}\bigg)\hat{i}+\bigg(\dfrac{-3+y}{2}\bigg)\hat{j}+\bigg(\dfrac{4+z}{2}\bigg)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\dfrac{x+2}{2}=\dfrac{3}{2}

\Rightarrow x=1

\dfrac{-3+y}{2}=\dfrac{7}{2}

\Rightarrow y=10

\dfrac{4+z}{2}=-\dfrac{9}{2}

\Rightarrow z=-13

Position vector of points D=\hat{i}+10\hat{j}-13\hat{k}

The vector equation of line BD passing through the points with position vectors \vec{a}(B) and \vec{b}(D) is \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})

Here,

\vec{a}=2\hat{i}-3\hat{j}+4\hat{k}

\vec{b}=\hat{i}+10\hat{j}-13\hat{k}

Vector equation of the required line is given by

\vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(\hat{i}+10\hat{j}-13\hat{k})     …..1

Here, \lambda is a parameter

Reducing (1) to cartesian form, we get

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(\hat{i}+10\hat{j}-13\hat{k})

\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(2-\lambda)\hat{i}+(-3+13\lambda)\hat{j}+(4-17\lambda)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\Rightarrow x=2-\lambda,~~y=-3+13\lambda~~z=4-17\lambda

\Rightarrow \dfrac{x-2}{-1}=\lambda,~\dfrac{y+3}{13}=\lambda,~\dfrac{z-4}{-17}=\lambda

Hence, the cartesian form of (1) is

\Rightarrow \dfrac{x-2}{-1}=\dfrac{y+3}{13}=\dfrac{z-4}{-17}=\lambda

 

 

Question 6

Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1,2,-1) and B (2,1,1) 

Sol :

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b} 

Here,

\vec{a}=\hat{i}+2\hat{j}-\hat{k}

\vec{b}=2\hat{i}+\hat{j}+\hat{k}

Vector equation of the required line is given by

\vec{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(2\hat{i}+\hat{j}+\hat{k})     …..1

Here, \lambda is a parameter

Reducing (1) to cartesian form, we get

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(2\hat{i}+\hat{j}+\hat{k})

\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(1+\lambda)\hat{i}+(2-\lambda)\hat{j}+(-1+2\lambda)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\Rightarrow x=1+\lambda,~~y=2-\lambda~~z=-1+2\lambda

\Rightarrow \dfrac{x-1}{1}=\lambda,~\dfrac{y-2}{-1}=\lambda,~\dfrac{z+1}{2}=\lambda

Hence, the cartesian form of (1) is

Rightarrow \dfrac{x-1}{1}=\dfrac{y-2}{-1}=\dfrac{z+1}{2}=\lambda

 

 

Question 7

Find in vector form as well as in cartesian form, the equation of the line passing through the points (1,2,3) and parallel to the vector \hat{i}+-2\hat{j}+3\hat{k} . Reduce the corresponding equation in cartesian form.

Sol :

 

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b} 

Here,

\vec{a}=\hat{i}+2\hat{j}+3\hat{k}

\vec{b}=\hat{i}-2\hat{j}+3\hat{k}

Vector equation of the required line is given by

\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})     …..1

Here, \lambda is a parameter

Reducing (1) to cartesian form, we get

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})

\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(1+\lambda)\hat{i}+(2-2\lambda)\hat{j}+(3+3\lambda)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\Rightarrow x=1+\lambda,~~y=2-2\lambda~~z=3+3\lambda

\Rightarrow \dfrac{x-1}{1}=\lambda,~\dfrac{y-2}{-2}=\lambda,~\dfrac{z-3}{3}=\lambda

Hence, the cartesian form of (1) is

\Rightarrow \dfrac{x-1}{1}=\dfrac{y-2}{-2}=\dfrac{z-3}{3}=\lambda

 

 

Question 8

Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \dfrac{x-3}{2}=\dfrac{y+1}{7}+\dfrac{z-2}{-3}

Sol :

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b} 

Here,

\vec{a}=\hat{i}+2\hat{j}+3\hat{k}

\vec{b}=\hat{i}-2\hat{j}+3\hat{k}

Vector equation of the required line is given by

\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-2\hat{j}+3\hat{k})     …..1

Here, \lambda is a parameter

 

 

Question 9

The cartesian equations of a line are \dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2} . Find a vector equation for the line.

Sol :

The cartesian equation of the given line is \dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2} .

It can be re-written as 

\dfrac{x-5}{3}=\dfrac{y-(-4)}{7}+\dfrac{z-6}{2}

Thus, the given line passes through the point having position vector

\vec{a}=5\hat{i}-4\hat{j}+6\hat{k} and is parallel to the vector \vec{b}=3\hat{i}+7\hat{j}+2\hat{k}

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b} 

Vector equation of the required line is given by

\vec{r}=(5\hat{i}-4\hat{j}+6\hat{k})+\lambda(3\hat{i}+7\hat{j}+2\hat{k})

Here, \lambda is a parameter

 

 

Question 10

Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are \dfrac{x-3}{1}=\dfrac{y-1}{2}=\dfrac{z+1}{-2} . Also, reduce the equation obtained in vector form.

Sol :

We know that the cartesian equation of a line passing through a point with position vector \vec{a} and parallel to the vector \vec{m} is \dfrac{x-x_1}{a}=\dfrac{y-y_2}{b}=\dfrac{z-z_3}{c}

Here,

\vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}

\vec{m}=a\hat{i}+b\hat{j}+c\hat{k}

Here, 

\vec{a}=\hat{i}-\hat{j}+2\hat{k}

\vec{b}=\hat{i}+2\hat{j}-2\hat{k}

Cartesian equation of the required line is 

\Rightarrow \dfrac{x-1}{1}=\dfrac{y-(-1)}{2}=\dfrac{z-2}{-2}

\Rightarrow \dfrac{x-1}{1}=\dfrac{y+1}{2}=\dfrac{z-2}{-2}

We know that the cartesian equation of a line passing through a point with position vector \vec{a} and parallel to \vec{m} is \vec{r}=\vec{a}+\lambda\vec{m} 

Here, the line is passing through the point (1,1,-2) and its direction ratios are proportional to 1,2,-2

Vector equation of the required line is 

\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\lambda(\hat{i}+2\hat{j}-2\hat{k})

 

 

Question 11

Find the direction cosines of the line \dfrac{4-x}{2}=\dfrac{y}{6}=\dfrac{1-z}{3} . Also, reduce it to vector form.

Sol :

The cartesian equation of the given line is 

\dfrac{4-x}{2}=\dfrac{y}{6}=\dfrac{1-z}{3}

It can be re-written as 

\dfrac{x-4}{-2}=\dfrac{y-0}{6}=\dfrac{z-1}{-3}

This shows that the given line passes through the point (4,0,1) and its direction ratios are proportional to -2 , 6 , -3 .

So , the direction ratios are 

\Rightarrow \dfrac{-2}{\sqrt{(-2)^2+(6)^2+(-3)^2}},~~\dfrac{6}{\sqrt{(-2)^2+(6)^2+(-3)^2}}~~\dfrac{-3}{\sqrt{(-2)^2+(6)^2+(-3)^2}}

\dfrac{-2}{7},~~\dfrac{6}{7}~~,\dfrac{-3}{7}

Thus,the given line passing through a point with position vector \vec{a}=4\hat{i}+\hat{k} and parallel to \vec{b}=-2\hat{i}+6\hat{j}-3\hat{k} is \vec{r}=\vec{a}+\lambda\vec{b} 

Here,

\vec{a}=4\hat{i}+\hat{k}

\vec{b}=-2\hat{i}+6\hat{j}-3\hat{k}

Vector equation of the required line is 

\vec{r}=(4\hat{i}+\hat{k})+\lambda(-2\hat{i}+6\hat{j}-3\hat{k})

Here, \lambda is a parameter.

 

 

Question 12

The cartesian equations of a line are x = ay,+ b , z = cy + d .Find its direction ratios and reduce it to vector form.

Sol :

The cartesian equation of the given line is 

x = ay + b ,  z = cy + d

It can be re-written as 

\dfrac{x-b}{a}=\dfrac{y-0}{1}=\dfrac{z-d}{c}

Thus, the given line passes through the point (b~,0~,d) and its direction ratios are proportional to a,1,c . It is also parallel to the vector \vec{b}=a\hat{i}+\hat{j}+c\hat{k}

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b} 

Vector equation of the required line is 

\vec{r}=(b\hat{i}+0\hat{j}+d\hat{k})+\lambda(a\hat{i}+j\hat{j}+c\hat{k})

Here, \lambda is a parameter.

 

 

Question 13

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 vector \hat{i}-\hat{j}+4\hat{k} and 2\hat{i}+\hat{j}+2\hat{k}. Also, find the cartesian equivalent of this equation. Sol : We know that the vector equation of a line passing through a point with position vector [latex]\vec{a} and parallel to the vector (\vec{b}-\vec{a}) is \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})

Here,

\vec{a}=\hat{i}-2\hat{j}-3\hat{k}

(\vec{b}-\vec{a})=(2\hat{i}+\hat{j}+2\hat{k})-(\hat{i}-\hat{j}+4\hat{k})

\Rightarrow \hat{i}+2\hat{j}-2\hat{k}

Vector equation of the required line is 

\vec{r}=(\hat{i}-2\hat{j}-3\hat{k})+\lambda(\hat{i}+2\hat{j}-2\hat{k})     ….1

Here, \lambda is a parameter

Reducing (1) to cartesian form , we get

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}-2\hat{j}-3\hat{k})+\lambda(\hat{i}+2\hat{j}-2\hat{k})

\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(1+\lambda)\hat{i}+(-2+2\lambda)\hat{j}+(-3-2\lambda)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\Rightarrow x=1+\lambda,~~y=-2+2\lambda~~z=-3-2\lambda

\Rightarrow \dfrac{x-1}{1}=\lambda,~\dfrac{y+2}{1}=\lambda,~\dfrac{z+3}{-2}=\lambda

Hence, the cartesian form of (1) is

\Rightarrow \dfrac{x-1}{1}=\dfrac{y+2}{1}=\dfrac{z+3}{-2}=\lambda

 

 

Question 14

Find the points on the line \dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2} at  a distance of 5 units from the point p (1,~3,~3)

Sol :

The coordinates of any point on the line \dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2} are given by 

\dfrac{x+2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{2}=\lambda

\Rightarrow x=3\lambda-2,~~y=2\lambda-1,~~z=2\lambda+3    ….1

Let the coordinates of the desired point be (3\lambda-2,~~2\lambda-1,~~2\lambda+3)[/lambda] The distance between this point and (1, 3, 3) is 5 units. [latex]\sqrt{(3\lambda-2-1)^2+(2\lambda-1-3)^2+(2\lambda+3-3)^2}=5

[Squaring both sides]

\Rightarrow (3\lambda-3)^2+(2\lambda-4)^2+(2\lambda)^2=25

\Rightarrow 17\lambda^2-34\lambda=0

\Rightarrow \lambda(\lambda-2)=0

\Rightarrow \lambda=0~~or~~2

Substituting the values of \lambda in (1), we get the coordinates of the desired point as (-2,~~-1,~~3) and (4,~~3,~~7)

 

 

Question 15

Show that the points whose position vectors are 2\hat{i}+3\hat{j},\hat{i}+2\hat{j}+3\hat{k}[latex]7\hat{i}-\hat{k}[/latex] collinear.

Sol :

Let the given points be P, Q and R and let their position vector be \vec{a}\vec{b} and \vec{c} respectively.

\vec{a}=-2\hat{i}+3\hat{j}

\vec{b}=\hat{i}+2\hat{j}+3\hat{k}

\vec{c}=7\hat{i}+9\hat{k}

Vector equation of line passing through P and Q is 

\Rightarrow \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})

\Rightarrow \vec{r}=(-2\hat{i}+3\hat{j})+\lambda{(\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j})}    …1

If points P,Q and R are collinear , the R must satisfy (1).

Replacing \vec{r} by \vec{c}=7\hat{i}+9\hat{k} in (1) , we get

\Rightarrow 7\hat{i}+9\hat{k}=(-2\hat{i}+3\hat{j})+\lambda{(\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j})}    …1

Comparing the coefficients \hat{i},\hat{j} and \hat{k} , we get

7=-2+3\lambda0=3-\lambda9=-3\lambda

\lambda=3

These three equations are consistent, i.e. they give the same value of \lambda . Hence , the given three points are collinear

 

 

Question 16

Find the cartesian and vector equations of a line which passes through the point (1,~2,~3) and is parallel to the line \dfrac{-x-2}{1}=\dfrac{y+3}{7}=\dfrac{2z-6}{3}

Sol :

we have 

\dfrac{-x-2}{1}=\dfrac{y+3}{7}=\dfrac{2z-6}{3}

It can be re-written as 

\dfrac{x+2}{-1}=\dfrac{y+3}{7}=\dfrac{z-3}{\dfrac{3}{2}}

\dfrac{x+2}{-2}=\dfrac{y+3}{14}=\dfrac{z-3}{3}

This shows that the given line passes through the point (-2,~-3,~3) and its direction ratios are proportional to -2 , 14 , 3 .

Thus , the parallel vector is \vec{b}=-2\hat{i}+14{j}+3\hat{k}

We know that the vector equation of a line passing through a point with position vector \vec{a} and parallel to the vector \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b}

Here,

\vec{a}=\hat{i}+2\hat{j}+3\hat{k}

\vec{b}=-2\hat{i}+14{j}+3\hat{k}

Vector equation of the required line is 

\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(-2\hat{i}+14{j}+3\hat{k})     …..1

Here, \lambda is a parameter

Reducing (1) to cartesian form, we get

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(-2\hat{i}+14{j}+3\hat{k})

\big[putting~~~\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}~~~in~(1)\big]

\Rightarrow x\hat{i}+y\hat{j}+z\hat{k}=(1-2\lambda)\hat{i}+(2+14\lambda)\hat{j}+(3+3\lambda)\hat{k}

Comparing the coefficients of \hat{i},~~\hat{j} and \hat{k},we get

\Rightarrow x=1-2\lambda,~~y=2+14\lambda~~z=3+3\lambda

\Rightarrow \dfrac{x-1}{-2}=\lambda,~\dfrac{y-2}{14}=\lambda,~\dfrac{z-3}{3}=\lambda

Hence, the cartesian form of (1) is

\Rightarrow \dfrac{x-1}{-2}=\dfrac{y-2}{14}=\dfrac{z-3}{3}=\lambda

 

 

Question 17

The cartesian equation of a line are 3x+1 = 6y-2 = 1-z . Find the fixed point through which it passes , its direction ratios and also its vector equation.

Sol :

The cartesian equation of the given line is 

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

It can be re-written as 

\dfrac{x+\dfrac{1}{3}}{\dfrac{1}{3}}=\dfrac{y-\dfrac{1}{3}}{\dfrac{1}{6}}=\dfrac{z-1}{-6}

Thus , the given line passes through the point (-\dfrac{1}{3},\dfrac{1}{3},1) and its direction ratios are proportional to (2,~1,-6). It is parallel to the vector \vec{b}=2\hat{i}+\hat{j}-6\hat{k}

We know that the vector equation of a line passes through a point with position vector \vec{b} is \vec{r}=\vec{a}+\lambda\vec{b}

Vector equation of the required line is 

\vec{r}=(-\dfrac{1}{3}\hat{i}+\dfrac{1}{3}\hat{j}+\hat{k})+\lambda(2\hat{i}+\hat{j}-6\hat{k})

Here , \lambda is a parameter.

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