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Class 12 RD Sharma Solutions – Chapter 28 The Straight Line in Space – Exercise 28.1 | Set 2

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  • Last Updated : 04 May, 2021
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Question 11. Find the direction cosines of the line\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3} . Also, reduce it to vector form

Solution:

Given:

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

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

x = -2λ + 4, y = 6λ, z = -3λ + 1

So,

x\hat{i}+y\hat{j}+z\hat{k}=(-2λ+4)\hat{i}+(6λ)\hat{j}+(-3λ+1)\hat{k}\\ \vec{r}=(4\hat{i}+\hat{k})+λ(-2\hat{i}+6\hat{j}-3\hat{k})

Direction ratios of the line are = -2, 6, -3

Direction cosines of the lines are,

\frac{a}{\sqrt{a^2+b^2+c^2}},\ \frac{b}{\sqrt{a^2+b^2+c^2}},\ \frac{c}{\sqrt{a^2+b^2+c^2}}\\ \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}}\\ \frac{-2}{7},\ \frac{6}{7},\ \frac{-3}{7}

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.

Solution:

x = ay + b

z = cy + d

\frac{x-b}{a}=\frac{y}{1}=\frac{z-d}{c}=\lambda

So, DR’s of line are (a, 1, c)

From above equation, we can write

x = aλ + b

y = λ

z = cλ + d

So vector equation of line is

x\hat{i}+y\hat{j}+z\hat{k}=(b\hat{i}+d\hat{k})+λ(a\hat{i}+\hat{j}+c\hat{k})

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 the position vector\hat{i}-\hat{j}+4\hat{k} and2\hat{i}+\hat{j}+2\hat{k} . Also, find the Cartesian equivalent of this equation.

Solution:

We know that, equation of a line passing through\vec{a} and parallel to vector\vec{b} is

\vec{r}=\vec{a}+λ\vec{b} ……. (i)

Here,

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

and, \vec{b} = line joining(\hat{i}-\hat{j}+4\hat{k}) and(2\hat{i}+\hat{j}+2\hat{k})

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

Equation of the line is

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

For Cartesian form of equation putx\hat{i}+y\hat{j}+z\hat{k}

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

Equating coefficients of\hat{i},\ \hat{j},\ \hat{k}

x = 1 + λ, y = -2 + 2λ, z = -3 – 2λ

\frac{x-1}{1}=λ,\ \frac{y+2}{2}=λ,\ \frac{z+3}{-2}=λ\\ \frac{x-1}{1}=\frac{y+2}{2}=\frac{z+3}{-2}

Question 14. Find the points on the line\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2} at a distance of 5 units from the points P(1, 3, 3).

Solution:

Given, line is\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda

General points Q on line is (3λ – 2, 2λ -1), 2λ + 3)

Distance of points P from Q =\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

PQ =\sqrt{(3λ-2-1)^2+(2λ-1-3)^2+(2λ+3-3)^2}

(5)2 = (3λ -3)2 + (2λ – 4) + (2λ)2

25 = 9λ2 + 9 – 18λ + 4λ2 + 16 – 16λ + 4λ2

17λ2 – 34λ = 0

17λ (λ – 2) = 0

λ = 0 or 2

So, points on the line are (3(0) – 2, 2(0) – 1, 2(0) + 3)

(3(2) – 2, 2(2) – 1, 2(2) + 3)

= (-2, -1, 3), (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} and7\hat{i}-\hat{k} are collinear.

Solution:

Let the given points are A,B.C with position vectors\vec{a},\ \vec{b},\ \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}-\hat{k}

We know that, equation of a line passing through\vec{a} and\vec{b} are

\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})\\ =(-2\hat{i}+3\hat{j})+λ((\hat{i}+2\hat{j}+3\hat{k})-(-2\hat{i}+3\hat{j}))\\ =(-2\hat{i}+3\hat{j})+\lambda(\hat{i}+2\hat{j}+3\hat{k}+2\hat{i}-3\hat{j})\\ \vec{r}=(-2\hat{i}+3\hat{j})+\lambda(3\hat{i}-\hat{j}+3\hat{k})\ \ \ \ .....(i)

If A, B, C are collinear then\vec{c} must satisfy equation (i)

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

Equation the coefficients of\vec{i},\ \vec{j},\ \vec{k}

-2 + 3 = 7 , λ = 3

3 – λ = 0 , λ = 3

3λ = -1 , λ =-\frac{1}{3}

Since, value of λ are not equal, so,

Given points are collinear.

Question 16. Find the Cartesian and vector equations of a line which passes through the points (1, 2, 3) and is parallel to the line\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}

Solution:

We know that, equation of a line passing through a point (x1, y1, z1) and having direction ratios proportional to a, b, c is

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\ \ \ \ .....(i)

Here,

(x1, y1, z1) = (1, 2, 3) and

Given line\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}

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

Its parallel to the required line, so

a = μ , b = 7μ, c =\frac{3}{2} μ

So, equation of required line using equation (i) is,

\frac{x-1}{-μ }=\frac{y-2}{7μ }=\frac{z-3}{\frac{3}{2}μ }\\ \frac{x-1}{-1}=\frac{y-2}{7}=\frac{z-3}{\frac{3}{2}}

Multiplying the denominators by 2

\frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3}=λ

x = -2λ + 1, y = 14λ + 2, z = 3λ + 3

So, vector form of the equation of required line,

x\hat{i}+y\hat{j}+z\hat{k}=(-2λ+1)\hat{i}+(14λ+2)\hat{j}+(3λ+3)\hat{k}\\ \vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+λ(-2\hat{i}+14\hat{j}+3\hat{k})

Question 17. The Cartesian equations 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.

Solution:

Given equation of line is,

3x + 1 = 6y -2 = 1 – z

Dividing all by 6

=\frac{3x+1}{6}=\frac{6y-2}{6}=\frac{1-z}{6}\\ =\frac{3x}{6}+\frac{1}{6}=\frac{6y}{6}-\frac{2}{6}=\frac{1}{6}-\frac{z}{6}\\ =\frac{1}{2}x+\frac{1}{6}=y-\frac{1}{3}=-\frac{z}{6}+\frac{1}{6}\\ =\frac{1}{2}\left(x+\frac{1}{3}\right)=1\left(y-\frac{1}{3}\right)=\frac{1}{6}(z-1)\\ =\frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}=λ

Comparing it with equation of line equation of line passing through (x,1 y1, z1) and the direction ratios a, b, c,

\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\\ (x_1,\ y_1,\ z_1)=\left(-\frac{1}{3},\ \frac{1}{3},\ 1\right)

a = 2, b = 1, -6

So, direction ratios of the line are -2, 1, -6

From equation (i)

x = \left(2λ-\frac{1}{3}\right),\ y=\left(λ+\frac{1}{3}\right),\ z=(-6λ+1)

So, vector equation of the given line is,

x\hat{i}+y\hat{j}+z\hat{k}=\left(2λ-\frac{1}{3}\right)\hat{i}+\left(λ+\frac{1}{3}\right)\hat{j}+(-6λ+1)\hat{k}\\ \vec{r}=\left(-\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}+\hat{k}\right)+λ(2\hat{i}+\hat{j}-6\hat{k})


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