RD Sharma Class 12 Ex 28.1 Solutions Chapter 28 The Straight Line in Space

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter28
Exercise28.1
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 28.1 Solutions Chapter 28 The Straight Line in Space

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Find the vector and Cartesian equation of the line through the points (5, 2, -4) and which is parallel to the vector3\hat{i}+2\hat{j}-8\hat{k}.

Solution:

As we know that the vector equation of a line is;

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

Thus, the Cartesian equation of a line is;

\frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{x-x_3}{a_3}

After applying the above formulas;

The vector equation of the line is;

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

The Cartesian equation of a line is;

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

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

Solution:

Given:

Here, the direction ratios of the line are;

(3 + 1, 4 – 0, 6 – 2) = (4, 4, 4)

Thus, the given line passes through

(-1, 0, 2)

As we know that the vector equation of a line is given as;

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

Thus, substitute values

Hence, we get

\vec{r}=\vec{a}+\lambda\vec{b}\\ \vec{r}=(-\vec{i}+0\vec{j}+2\vec{k})+\lambda(4\vec{i}+4\vec{j}+4\vec{k})

Therefore,

Vector equation of the line is;

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

Question 3. Fine the vector equation of a line which is parallel to the vector2\hat{i}-\hat{j}+3\hat{k} and which passes through the point (5, -2, 4), Also, reduce it to Cartesian form.

Solution:

Consider,

The vector equation of line passing through a fixed point vector a and parallel to vector b is shown as;

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

Here, λ is scalar

\vec{b}=2\hat{i}-\hat{j}+3\hat{k} and\vec{a}=5\hat{i}-2\hat{j}+4\hat{k}

The equation of the required line is;

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

Now substitute the value of r here

\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}

Thus, we get

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

Now compare the coefficients of vector

x = 5 + 2λ,y = -2 – λ,z = 4 + 3λ

After equating to λ,

We will have

\frac{x-5}{2}=λ ,\ \ \frac{y+2}{-0}=λ ,\ \ \frac{z-4}{3}=λ

Therefore,

The Cartesian form of equation of the line is;

\frac{x-5}{2} =\ \ \frac{y+2}{-0} =\ \ \frac{z-4}{3}

Question 4. A line passing through the point with position vector2\hat{i}-3\hat{j}+4\hat{k} and is in the direction of3\hat{i}+4\hat{j}-5\hat{k} . Find equations of the line in vector and Cartesian form.

Solution:

Consider,

The vector equation of line passing through a fixed point vector a and parallel to vector b is shown as;

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

Here, λ is scalar

\vec{a}=2\hat{i}-3\hat{j}+4\hat{k} and\vec{b}=3\hat{i}+4\hat{j}-5\hat{k}

The equation of the required line is;

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

Now substitute the value of r here

\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}

Thus, we get

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

Now compare the coefficients of vector

x = 2 + 3λ,y = -3 + 4λ,z = 4 – 5λ

After equating to λ,

We will have

\frac{x-2}{3}=λ ,\ \ \frac{y+3}{4}=λ ,\ \ \frac{z-4}{-5}=λ

Therefore,

The Cartesian form of equation of the line is;

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

Question 5. ABCD is a parallelogram. The position vectors of the points A, B and 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.

Solution:

Given: ABCD is a parallelogram.

Consider: AC and BD bisects each other at point O.

Thus,

Position vector of point O =\frac{\vec{a}+\vec{c}}{2}\\ =\frac{(4\hat{i}+5\hat{j}-10\hat{k})+(-\hat{i}+2\hat{j}+\hat{k})}{2}\\ =\frac{3\hat{i}+7\hat{j}-9\hat{k}}{2}

Now, Consider position vector of point O and B are represented by

\vec{o} and\vec{b}

Thus,

Equation of the line BD is the line passing through O and B is given by

\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) [Since equation of the line passing through two points\vec{a} and\vec{b} ]

\vec{r}=\vec{a}+\lambda(\vec{o}-\vec{b})\\ (2\hat{i}-3\hat{j}+4\hat{k})+\lambda\left(\frac{3\hat{i}+7\hat{j}-9\hat{k}}{2}-2\hat{i}-3\hat{j}+4\hat{k}\right)\\ \vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(3\hat{i}+7\hat{j}-9\hat{k}-4\hat{i}+6\hat{j}-8\hat{k})\\ \vec{r}=(2\hat{i}-3\hat{j}+4\hat{k})+\lambda(-\hat{i}+13\hat{j}-17\hat{k})

Now, compare the coefficients of vector i, j, R

x = 2 – λ, y = -3 – 13λ, z = 4 – 17λ

After equating to λ,

We will have

\frac{x-2}{-1}=λ ,\ \ \frac{y+3}{13}=λ ,\ \ \frac{z-4}{-17}=λ

Therefore,

The Cartesian form of equation of the line is;

\frac{x-2}{-1} =\ \ \frac{y+3}{13} =\ \ \frac{z-4}{-17}

Question 6. Find the vector form as well as in Cartesian form, the equation of line passing through the points A(1, 2, -1) and B(2, 1, 1).

Solution:

We know that, equation of line passing though two points (x1, y1 ,z1) and (x2, y2, z2) is

\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\ \ \ \ \ \ \ ....(i)\\

Here,

(x1, y1, z1) = A(1, 2, -1)

(x2, y2 ,z2) = B(2, 1, 1)

Using equation (i), equation of line AB,

\frac{x-1}{2-1}=\frac{y-2}{1-2}=\frac{z+1}{1+1}\\ \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2}=\lambda\ (assume)

x = λ + 1, y = -λ + 2, z = 2λ – 1

Vector form of equation of line AB is,

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

Question 7. Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector\hat{i}-2\hat{j}+3\hat{k}. Reduce the corresponding equation in Cartesian form.

Solution:

We know that vector equation of a line passing through\vec{a} and parallel to the vector\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}+3\hat{k}

So, required vector equation of line is,

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

Now,

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

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

x = 1 + λ, y = 2 – 2λ, z = 3 + 3λ

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

So, required equation of line is Cartesian form,

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

Question 8. Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-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) = (2, -1, 1) and

Given line\frac{x-3}{2}=\frac{y+1}{7}=\frac{z-2}{-3} is parallel to required line.

a = 2μ, b = 7μ, c = -3μ

So, equation of required line using equation (i)

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

x = 2λ + 2, y = 7λ – 1, z = -3λ + 1

So,

x\hat{i}+y\hat{j}+z\hat{k}=(2λ+2)\hat{i}+(7λ-1)\hat{j}+(-3λ+1)\hat{k}\\ \vec{r}=(2\hat{i}-\hat{j}+\hat{k})+λ(2\hat{i}+7\hat{j}-3\hat{k})

Question 9. The Cartesian equation of a line is\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} . Write its vector form

Solution:

The Cartesian equation of the line is

\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} ….(i)

The given line passes through the point (5, -4, 6). The position vector of this point is

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

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

This means that the line is in the direction of vector,

\vec{b}=3\hat{i}+7\hat{j}+2\hat{k}

It is known that the line through position vector\vec{a} and in the direction of the vector\vec{b} is given by the equation,

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

Question 10. Find the Cartesian equation of a line passing through (1, -1, 2) and parallel to the line whose equations are \frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}. Also, reduce the equation obtained in vector form.

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, -1, 2) and

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

So,

a = μ, b = 2μ, c = -2μ

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

\frac{x-1}{μ }=\frac{y+1}{2μ }=\frac{z-2}{-2μ }\\ \frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}=\lambda

x = λ + 1, y = 2λ – 1, z = -2λ +2

So,

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

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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