Peer review of ML Aggarwal Class 12 Solutions Chapter 2 Three Dimensional Geometry Ex 2.1 can encourage collaborative learning.

## ML Aggarwal Class 12 Maths Solutions Section B Chapter 2 Three Dimensional Geometry Ex 2.1

Very short answer type questions (1 to 8) :

Question 1.

(i) Find the distance of the point P(2, – 3, 4) from the x-axis.

(ii) Find the distance of the point Q(5, – 4, 12) from they-axis.

(iii) Find the distance of the point P (5, – 6, 3) from the XOY plane.

(iv) Find the distance of the point Q (2, – 4, 6) from the ZOX plane.

Solutions:

(i) Let M be the foot of! drawn from point P (2, – 3, 4) to then.x-axis.

Then coordinates of M are (2, 0, 0)

∴ |PM| = Required distance

= \(\sqrt{(2-2)^2+(-3-0)^2+(4-0)^2}\)

= \(\sqrt{9+16}\)

= 5 units

(ii) Let M be the foot of ⊥ drawn from point Q (5, – 4, 12) to the y-axis.

Then the coordinates of M are (0, – 4, 0).

∴ required distance = |QM|

= \(\sqrt{(5-0)^2+(-4+4)^2+(12-0)^2}\)

= \(\sqrt{25+144}\)

= 13 units

(iii) Let M be the foot of ⊥ drawn from point P(5, – 6, 3) to XOY plane

i.e. z = 0 then coordinates of M are (5, – 6, 0).

∴ required distance = |PM|

= \(\sqrt{(5-5)^2+(-6+6)^2+(3-0)^2}\)

= 3 units

(iv) Let M be the foot of ⊥ drawn from the point Q (2, – 4, 6) to ZOX plane i.e. y = 0.

Then coordinates of M are (2, 0, 6)

∴ required distance = |QM|

= \(\sqrt{(2-2)^2+(-4-0)^2+(6-6)^2}\)

= 4 units

Question 2.

(i) If a line makes angles 90°, 60° and 30° with the positive directions of x, y and z- axes respectively, find its direction cosines. (NCERT)

(ii) Write the direction cosines of z-axis.

Solutions:

(i) Since the line makes anglesof 90°, 60° and 30° with positive direction of x, y and z-axis.

∴ direction cosines of the line are cos 90°, cos 60°, cos 30° i.e. 0, \(\frac{1}{2}\), \(

(ii) Now z-axis made an angle of 90°, 90° and 0° with xaxis, y-axis and z-axis.

Thus direction cosines of z-axis are

< cos 90°, cos 90°, cos 0° > i.e.. < 0, 0, 1>

Question 3.

(i) Can the numbers [latex]\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\) be the direction cosines of a line ?

(ii) Can a line have direction angles 45°, 60°, 120°?

(iii) Prove that 1, 1, 1 cannot be direction cosines of a straight line.

Solutions:

(i) Here l^{2} + m^{2} + n^{2}

= \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2\)

= \(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)

= \(\frac{3}{2}\) ≠ 1

∴ \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\) can’t be the direction cosines of a line.

(ii) Here l = cos α

= cos 45°

= \(\frac{1}{\sqrt{2}}\) ;

m = cos 60° = \(\frac{1}{2}\)

and n = cos 120°

= cos (180 – 60)

= – cos 60°

= – \(\frac{1}{2}\)

Here, l^{2} + m^{2} + n^{2} = \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)^2\)

= \(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\)

= 1

∴ 45°, 60°, 120° can be the direction angles of a line.

(iii) Since l^{2} + m^{2} + n^{2} = 1^{2} + 1^{2} + 1^{2}

= 3 ≠ 1

Question 3 (old).

(ii) Write the direction cosines of a line equa1ly inclined to the three coordinate axes.

Solution:

Let α, β, γ be the angles made by line with coordinate axis.

Let < l, m, n > be the direction cosines of given line

and line is equally inclined to coordinate axes.

∴ l = cos α; m = cos α; n = cos α [∵ α = β = γ]

Also, l^{2} + m^{2} + n^{2} = 1

⇒ cos^{2} α + cos^{2} α + cos^{2} α = 1

⇒ cos α = ± \(\frac{1}{\sqrt{3}}\)

Hence, the direction cosine’s of given line are < \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\) >

or < \(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\) >

Question 4.

If the direction cosines of a line are < \(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\) >, then find values(s) of c.

Solution:

Since the direction cosines of line are < \(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\) >

∴ l = m = n = \(\frac{1}{c}\)

and l^{2} + m^{2} + n^{2} = 1

⇒ \(\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}\) = 1

⇒ \(\frac{3}{c^2}\) = 1

⇒ c^{2} = 3

⇒ c = ± √3

Question 5.

(i) If a line makes angles 90°, 60° and θ with x, y and z-axis respectively, where θ is acute, then find θ.

(ii) If a line makes angles of 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.

(iii) A line makes an angle of with each of X-axis and Y-axis. What angle does it makes with Z-axis.

(iv) If a straight line makes angles of 45° and 120° with the positive directions of x and z-axes respectively, then find the acute angle which it makes with the positive direction of y-axis.

(v) If the direction cosines of a line are \(\frac{2}{3},-\frac{1}{3},-\frac{2}{3}\), then find its direction ratios.

Solution:

(i) We know that, ifa line makes an angles α, β, γ with coordinate axes.

Then cos^{2} α + cos^{2} β + cos^{2} γ = 1

⇒ cos^{2} 90° + cos^{2} 60°+ cos^{2} θ = 1

⇒ 0 + (\(\frac{1}{2}\))^{2} + cos^{2} θ = 1

⇒ cos^{2} θ = \(\frac{3}{4}\)

⇒ cos θ = ± \(\frac{\sqrt{3}}{2}\)

When cos θ = \(\frac{\sqrt{3}}{2}\)

⇒ θ = \(\frac{\pi}{6}\)

When cos θ = – \(\frac{\sqrt{3}}{2}\)

⇒ cos θ = cos (π – \(\frac{\pi}{6}\))

⇒ θ = \(\frac{5\pi}{6}\)

Hence, the required angles made by line with z-axis are \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).

Thus the required acute angle be \(\frac{\pi}{6}\).

(ii) Let θ be the angle made by line with positive direction of z-axis.

Thus direction cosines of line are

i.e. if < l, m, n> are direction cosines of line

Then l = cos 90° = 0;

m = cos 60° = \(\frac{1}{2}\) ;

n = cos θ

Also, l^{2} + m^{2} + n^{2} = 1

⇒ 0^{2} + (\(\frac{1}{2}\))^{2} + cos^{2} θ = 1

⇒ cos^{2} θ = \(\frac{3}{4}\)

= \(\left(\frac{\sqrt{3}}{2}\right)^2\)

= cos^{2} \(\frac{\pi}{6}\)

⇒ θ = nπ ± \(\frac{\pi}{6}\) ∀ n ∈I

since 0 ≤ θ ≤ π

∴ θ = \(\frac{\pi}{6}\), π – \(\frac{\pi}{6}\) = 30°, 150°

(iii) Let < l, m, n > be the direction cosines of line.

Also direction angles of line are ;

cos \(\frac{\pi}{4}\), cos \(\frac{\pi}{4}\), cos θ

where θ be the angle made by line with Z-axis.

∴ l = cos \(\frac{\pi}{4}\) ;

m = cos \(\frac{\pi}{4}\) ;

n = cos θ

⇒ l^{2} + m^{2} + n^{2} = 1

⇒ cos^{2} \(\frac{\pi}{4}\) + cos^{2} \(\frac{\pi}{4}\) + cos^{2} θ = 1

⇒ \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2\) + cos^{2} θ = 1

⇒ \(\frac{1}{2}+\frac{1}{2}\) + cos^{2} θ = 1

⇒ cos^{2} θ = 0

⇒ cos θ = 0

⇒ θ = \(\frac{\pi}{2}\)

(iv) Let θ be the angle between the line and positive direction of y-axis.

Thus direction cosines of line are < cos 45°, cos θ, cos 120°> if < l, m, n > are the direction cosines of line.

then l = cos 45° = \(\frac{1}{\sqrt{2}}\) ;

n = cos 120° = cos (180° – 60°) = – cos 60°

i.e. m = – \(\frac{1}{2}\)

and n = cos θ

since, l^{2} + m^{2} + n^{2} = 1

⇒ \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(-\frac{1}{2}\right)^2\) + cos^{2} θ = 1

⇒ \(\frac{1}{2}+\frac{1}{4}\) + cos^{2} θ = 1

⇒ cos^{2} θ = \(\frac{1}{4}\)

= (\(\frac{1}{2}\))^{2}

= cos^{2} \(\frac{\pi}{3}\)

∴ θ = nπ ± \(\frac{\pi}{3}\) ∀ n ∈1

Thus θ = \(\frac{\pi}{3}\), π – \(\frac{\pi}{3}\) i.e. 60°, 120° [∵ 0 ≤ θ ≤ π]

∴ required acute angle be 60°.

(v) Given direction cosines of given line are < \(\frac{2}{3},-\frac{1}{3},-\frac{2}{3}\) >

∴ l = \(\frac{2}{3}\) ;

m = – \(\frac{1}{3}\) ;

n = – \(\frac{2}{3}\)

where < l, m, n> are the D’cosines of given line.

Let < a, b, c > be the direction ratios of given line.

∴ a = λl = \(\frac{2}{3}\) λ ;

b = λm = – \(\frac{\lambda}{3}\) ;

c = λn = – \(\frac{2}{3}\) λ

∴ direction ratios of given line are < + \(\frac{2}{3}\) r, – \(\frac{\lambda}{3}\), – \(\frac{2}{3}\) λ >

i.e. < 2λ, – λ, – 2λ > where λ ≠ 0

Question 6.

(,) If a line has direction ratios 2, – 1, – 2, determine its direction cosines.

(ii) If direction numbers of a line are < 2, – 6, 3 >, find its direction cosines.

Solution:

(î) Given D’ratios of line are < 2, – 1, – 2 >

∴ D’cosines of line are

< \(\frac{2}{\sqrt{2^2+(-1)^2+(-2)^2}}, \frac{-1}{\sqrt{2^2+(-1)^2+(-2)^2}}, \frac{-2}{\sqrt{2^2+(-1)^2+(-2)^2}}\) >

i.e. < \(\frac{2}{3}, \frac{-1}{3}, \frac{-2}{3}\) >

(ii) Given direction ratios of given line are < 2, – 6, 3 >

∴ direction cosines of given line are :

< \(\frac{2}{\sqrt{2^2+(-6)^2+3^2}}, \frac{-6}{\sqrt{2^2+(-6)^2+3^2}}, \frac{3}{\sqrt{2^2+(-6)^2+3^2}}\) >

i.e. < \(\frac{2}{7},-\frac{6}{7}, \frac{3}{7}\) >

Question 7.

(i) Write the direction cosines of the line joining the points (1, 0, 0) and (0, 1, 1).

(ii) Find the direction cosines of the line passing through the points P(2, 3, 5) and Q (- 1, 2, 4). (NCERT Exemplar)

Solution:

(i) We know that the Direction ratios of the line joining

P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) are <x_{2}, – x_{1}, y_{2} – y_{1}, z_{2} – z_{1}>

∴ D’ ratios of the line are < 0 – 1, 1 – 0, 1 – 0 > i.e. <- 1, 1, 1 >

∴ Direction cosines of the line are < \(\frac{-1}{\sqrt{1^2+1^2+1^2}}, \frac{1}{\sqrt{1^2+1^2+1^2}}, \frac{1}{\sqrt{1^2+1^2+1^2}}\) >

i.e. < \(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\) >

(ii) D’ ratios of the line passing through the points p (2, 3, 5) and Q (- 1, 2, 4) be

<- 1 – 2, 2 – 3, 4 – 5 > i.e. < – 3, – 1, – 1 > i.e. < 3, 1, 1 >

∴ D’ cosines of given line be : < \(\frac{3}{\sqrt{3^2+1^2+1^2}}, \frac{1}{\sqrt{3^2+1^2+1^2}}, \frac{1}{\sqrt{3^2+1^2+1^2}}\) >

i.e. < \(\frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}}, \frac{1}{\sqrt{11}}\) >

Question 8.

(i) Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1) and (4, 3, – 1). (NCERT)

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

Solution:

(i) Direction numbers of line joining O (0, 0, 0) and A (2, 1, 1) are < 2, 1, 1 >

and D’numbers of line joining B (3, 5, – 1) and C (4, 3, – 1) are < 4 – 3, 3 – 5, 0>

i.e. < 1, – 2, 0 >

Here a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 2(1) + 1 (- 2) + 1 – 0 = 0

∴ both lines are perpendicular.

(ii) Direction ratios of the line (1) through the points (1, – 1, 2) and (3, 4, – 2) are proportional to,

< 3 – 1, 4 – (- 1), – 2 – 2 > i.e. < 2, 5, – 4 >

Here, a_{1} = 2, b_{1} = 5, c_{1} = – 4

Now direction ratios of line (2) through the points (0, 3, 2) and (3, 5, 6) are proportional to

< 3 – 0, 5 – 3, 6 – 2 > i.e. < 3, 2, 4 >

Here a_{2} = 3, b_{2} = 2, c_{2} = 4

Here, a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 2 (3) + 5 (2) + (- 4) 4 = 0

∴ line (1) is ⊥to line (2).

Question 9.

(i) For what value of p will the line through (4, 1,2) and (5, p, 0) be perpendicular to the line through (2, 1, 1) and (3, 3, – 1).

(ii) Find the values of p and q so that the line joining the points (7, p, 2) and (q, – 2, 5) may be parallel to the line joining the points (2, – 3, 5) and (- 6, – 15, 11).

Solution:

(i) Direction ratios of line AB are < 5 – 4, p – 1,0 – 2 > i.e. < 1, p – 1, – 2 >

Direction ratios of line CD are < 3 – 2, 3 – 1, – 1 – 1 > i.e. < 1, 2, – 2 >

Since line AB is perpendicular to line CD

∴ a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

⇒ 1.1 + (p – 1) 2 + (- 2) (- 2) = 0

⇒ 1 + 2p – 2 + 4 = 0

⇒ 2p + 3 = 0

⇒ p = – \(\frac{3}{2}\)

(ii) We know that, two lines with direction ratios < a_{1}, b_{1}, c_{1} > and < a_{2}, b_{2}, c_{2} > are parallel iff \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

D’ratios of line AB are < q – 7, – 2 – p, 5 – 2 > i.e. < q – 7, – 2 – p, 3 >

D’ ratios of line CD are < – 6 – 2, – 15 + 3, 11 – 5 > i.e. < – 8, – 12, 6 >

∴ \(\frac{q-7}{-8}=\frac{-2-p}{-12}=\frac{3}{6}\)

⇒ \(\frac{7-q}{8}=\frac{p+2}{12}=\frac{1}{2}\)

∴ p + 2 = 6

⇒ p = 4

and 7 – q = 4

⇒ q = 3

Question 10.

Show that the points (2, 3, 4), (- 1, – 2, 1) and (5, 8, 7) are collinear. (NCERT)

Solution:

Let A, B, C are given points whose position vectors are (2, 3, 4), (-1, -2, 1) and (5, 8, 7)

∴ direction ratios of line AB are proportional to < – 1 – 2, – 2 – 3, 1 – 4 > i.e. <- 3, – 5, – 3 >

direction ratios of line BC are proportional to < 5 – (- 1), 8 – (- 2), 7 – 1 > i.e. < 6, 10, 6 >

also \(\frac{-3}{6}=\frac{-5}{10}=\frac{-3}{6}\)

∴ direction ratios of both lines are proportional and B be a point common to both lines.

Thus A, B and C are collinear.

Question 11.

Find the direction ratios and the direction cosines of the line joining the point (5, 4, – 3) and (3, – 2, 6), the angle between the line and X-axis taken acute.

Solution:

D’ratio’s of line joining A (5, 4,- 3) and B (3,- 2,6) are < 3 – 5, 2 – 4, 6 + 3 >

i.e. <- 2, – 6, 9>

∴ D cosines of line are < \(\frac{-2}{\sqrt{(-2)^2+(-6)^2+9^2}}, \frac{-6}{\sqrt{(-2)^2+(-6)^2+9^2}}, \frac{9}{\sqrt{(-2)^2+(-6)^2+9^2}}\) >.

i.e. < \(\frac{-2}{11},-\frac{6}{11}, \frac{9}{11}\) >

Let θ be the angle made by line with x-axis whose d’ cosines are < 1, 0, 0 >

∴ cos θ = \(\frac{|(-2)(1)-6(0)+9(0)|}{\sqrt{(-2)^2+(-6)^2+9^2} \cdot 1}\)

= \(\frac{|-2|}{11}=\frac{2}{11}\)

[∵ θ being acute angle]

Question 12.

(i) Find the angle between the lines whose direction cosines are < \(\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\) > and < \(-\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\) >.

(ii) Find the angle between the lines whose direction numbers are < 1, 2, 1 >, < 2, – 3, 4>

(iii) Find the angle between the lines whose direction ratios are < 1, 1, 2 > and <√3 – 1, – √3, – 1, 4>.

Solution:

(i) Here l_{1} = \(\frac{\sqrt{3}}{2}\) ;

m_{1} = \(\frac{1}{4}\) ;

n_{1} = \(\frac{\sqrt{3}}{4}\)

Let θ be the angle between two given lines

∴ cos θ = |l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}|

= \(\left|\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{4} \times \frac{1}{4}+\frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{4}\right|\)

= \(\left|-\frac{3}{4}+\frac{1}{16}+\frac{3}{16}\right|\)

= |- \(\frac{1}{2}\)|

⇒ cos θ = \(\frac{1}{2}\)

⇒ θ = \(\frac{\pi}{3}\)

(ii) Here a_{1} = 1 ;

b_{1} = 2 ;

c_{1} = 1 ;

a_{2} = 2

b_{2} = – 3

c_{2} = 4

Let θ be the angle between two given lines.

∴ cos θ =|l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}

= \(\left|\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{4} \times \frac{1}{4}+\frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{4}\right|\)

= \(\frac{1(2)+2}{\sqrt{1^2+2^2+1^2}} \frac{(-3)+1(4)}{\sqrt{2^2+(-3)^2+4^2}}\)

⇒ cos θ = 0

⇒ θ = 90°

(iii) Here, a_{1} = 1;

b_{1} = 1;

c_{1} = 2;

a_{2} = √3 – 1 ;

b_{2} = – √3 – 1 ;

c_{2} = 4

Let θ be the required angle between two given lines.

Then cos θ = \(\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)

= \(\frac{1(\sqrt{3}-1)+1(-\sqrt{3}-1)+2(4)}{\sqrt{1^2+1^2+2^2} \sqrt{(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2+4^2}}\)

= \(\frac{6}{\sqrt{6} \sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}+16}}\)

= \(\frac{6}{\sqrt{6} \sqrt{24}}=\frac{6}{\sqrt{144}}=\frac{6}{12}=\frac{1}{2}\)

θ = \(\frac{\pi}{3}\)

Question 13.

Show that the lines with direction cosines < \(\frac{12}{13},-\frac{3}{13},-\frac{4}{13}\) >, < \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\) > and < \(\frac{3}{13},-\frac{4}{13}, \frac{12}{13}\) > are mutually perpendicualr.

Solution:

The Direction cosines of three lines are < \(\frac{12}{13},-\frac{3}{13},-\frac{4}{13}\) > ;

< \(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\) >

and < \(\frac{3}{13},-\frac{4}{13}, \frac{12}{13}\) >

Now we know that lines with D’ cosines

< l_{1}, m_{1}, n_{1} > and < l_{2}, m_{2}, n_{2} >

if l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 0

Thus, lines (1) and (2) are ⊥

if \(\left(\frac{12}{13}\right) \frac{4}{13}-\frac{3}{13} \cdot \frac{12}{13}-\frac{4}{13} \cdot \frac{3}{13}\) = 0

i.e. if 0 = 0 which is true

lines (2) and (3) are ⊥ if

\(\left(\frac{4}{13}\right) \frac{3}{13}+\left(\frac{12}{13}\right)\left(-\frac{4}{13}\right)+\left(\frac{3}{13}\right)\left(\frac{12}{13}\right)\) = 0

i.e. if 0 = 0 which is true

lines (1) and (3) are ⊥ if

\(\left(\frac{12}{13}\right)\left(\frac{3}{13}\right)+\left(\frac{-3}{13}\right)\left(\frac{-4}{13}\right)+\left(\frac{-4}{13}\right)\left(\frac{12}{13}\right)\)

i.e. if 0 = 0 which is true

Hence the given lines are mutually perpendicular.

Question 14.

If A, B are the points (2, 1, – 2), (3, – 4, 5), then find the angle that OA makes with OB, where O is the origin.

Solution:

D’ratios of line OA are < 2, 1, – 2 >

D’ratios of line OB are < 3, – 4, 5 >

Let θ be the acute angle between OA and OB

Then cos θ = \(\frac{|2(3)+1(-4)-2(5)|}{\sqrt{2^2+1^2+(-2)^2} \sqrt{3^2+(-4)^2+5^2}}\)

= \(\frac{8}{3 \times 5 \sqrt{2}}=\frac{4 \sqrt{2}}{15}\)

∴ θ = cos^{-1} \(\left(\frac{4 \sqrt{2}}{15}\right)\)

Question 15.

If the coordinates of the points A, B, C and D are (1, 2, 3), (4, 5, 7), (- 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD. (NCERT)

Solution:

Direction ratios of line AB are proportional to < 4 – 1, 5 – 2, 7 – 3 > i.e. < 3, 3, 4 >

and the direction ratio of line CD are proportional to < 2 – (- 4), 9 – 3, 2 – ( – 6) >

i.e. < 6, 6, 8 >

i.e. < 3, 3, 4 >

Let θ be the angle between lines AB and CD.

Here a_{1} = 3 ;

b_{1} = 3 ;

c_{1} = 4

and a_{2} = 3 ;

b_{2} = 3 ;

c_{2} = 4

∴ cos θ = \(\frac{3(3)+3(3)+4(4)}{\sqrt{3^2+3^2+4^2} \sqrt{3^2+3^2+4^2}}\)

= \(\frac{34}{34}\) = 1

⇒ θ = 0.

Question 16.

Show that the points A (4, 7, 8), B (2, 3, 4), C (- 1, – 2, 1) and D (1, 2, 5), taken in order, are the vertices of a parallelogram.

Solution:

Direction ratios of line AB are

< 2 – 4, 3 – 7, 4 – 8 > i.e.< – 2, – 4, – 4 >

D’ ratios of line CD are

< 1 + 1, 2 + 2, 5 – 1 > i.e. < 2, 4, 4 >

∴ D’ ratios of lines AB and DC are proportional.

∴ lines AB and DC are parallel.

Also D’ ratios of line AD are ;

< 1 – 4, 2 – 7, 5 – 8 > i.e. < – 3, – 5, – 3 >

D’ ratios of line BC are

< – 1 – 2, – 2 – 3, 1 – 4 > i.e. <- 3, – 5, – 3 >

∴ d’ ratios of both lines AD and BC are proportional.

Thus, lines AD and BC are parallel.

Hence ABCD be a parallelogram.

Question 17.

Find the direction cosines of a line which is perpendicular to the two lines formed by joining A (2, 3, – 4) to B (- 3, 3, – 2) and C (- 1, 4, 2) to D (3, 5, 1).

Solution:

The d’ratios of line AB are < – 3 – 2, 3 – 3, – 2 + 4> i.e. < – 5, 0, 2 >

The direction ratios of line CD are < 3 + 1, 5 – 4, 1 – 2 > i.e. < 4, 1, – 1 >

Let < a, b, c > be the direction ratios of the required line.

Since it is given that, required line is ⊥ to AB and line CD.

∴ – 5a + 0b + 2c = 0 ;

4b + b – c = 0 [∵ a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0]

\(\frac{a}{0-2}=\frac{b}{8-5}=\frac{c}{-5}\) i.e. \(\frac{a}{-2}=\frac{b}{3}=\frac{c}{-5}\)

∴D’ ratios of line are <- 2, 3, – 5 >

Thus. D’ cosines of line are < \(\frac{-2}{\sqrt{(-2)^2+3^2+5^2}}, \frac{3}{\sqrt{(-2)^2+3^2+(-5)^2}}, \frac{-5}{\sqrt{(-2)^2+3^2+(-5)^2}}\) >

i.e. < \(\frac{-2}{\sqrt{38}}, \frac{3}{\sqrt{38}}, \frac{-5}{\sqrt{38}}\) >

Question 18.

Find the angle between the lines whose direction cosines satisfy the relations l + m + n = 0 and l^{2} = m^{2} + n^{2}.

Solution:

The equations giving direction cosines of the two lines are;

l + m + n = 0 …………………..(1)

l^{2} – m^{2} – n^{2} = 0 ………………(2)

From (1) ;

l = – m – n

putting this value of l in eqn. (2); we get

⇒ (- m – n)^{2} – m^{2} – n^{2} = 0

⇒ m^{2} + n^{2} + 2mn – m^{2} – n^{2} = 0

⇒ 2mn = 0

⇒ mn = 0

either m = 0 or n = 0

∴ from (1);

l = – n

⇒ l = – m [using (1)]

Thus direction cosines of two lines are ;

< – n, 0, n > and < – m, m, 0 >

i.e., < – 1, 0, 1> and < – 1, 1, 0 >

Let θ be the angle between them

Then cos θ = \(\frac{|-1(-1)+0 \times 1+1 \times 0|}{\sqrt{(-1)^2+0^2+1^2} \sqrt{(-1)^2+1^2+0^2}}\)

= \(\frac{1}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\)

⇒ θ = \(\frac{\pi}{3}\)

Question 19.

(i) Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles. (NCERT Exemplar)

(ii) Find the angle between the lines whose direction cosines are given by the equations 3l + m + 5n = 0 and 6m – 2nl + 5lm = 0. (NCERT Exemplar)

Solution:

(i) Given 2l + 2m – n = 0 ……………….(1)

mn + ln + lm = 0

From (1);

n = 2l + 2m

putting in eqn. (2); we get

(l + m) (2l + 2m) + lm = 0

⇒ 2l^{2} + 5lm + 2m^{2} = 0

⇒ (2l + m) (2m + l) = 0

⇒ l = – \(\frac{m}{2}\), – 2m

When l = – \(\frac{m}{2}\)

∴ from (1) ; we have

n = 2 (- \(\frac{m}{2}\)) + 2m = m ;

When l = – 2m

∴ from (1) ;

n = 2 (- 2m) + 2m = – 2m

∴ The direction ratios of the lines are proportional to < – \(\frac{m}{2}\), m, m > i.e. < – 1, 2, 2 >

and < – 2m, m, – 2m > i.e. < – 2, 1, – 2 >

Thus the vector || to these lines are \(\vec{a}=-\hat{i}+2 \hat{j}+2 \hat{k}\) and \(\vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k}\)

If θ be the angle between the lines θ be the angle between \(\vec{a} \text { and } \vec{b}\).

∴ cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)

= \(\frac{-1(-2)+2(1)+2(-2)}{\sqrt{1+4+4} \sqrt{4+1+4}}\)

= \(\frac{0}{9}\) = 0

θ = \(\frac{\pi}{2}\)

(ii) The equations giving direction cosines of the two lines are

3l + 5m + 5n = 0 ……………(1)

and 6mn – 2nl + 5lm = 0 ……………(2)

From (1) ;

m = – 3l – 5n

∴ from (2) ;

(6n + 5l) (- 3l – 5n) – 2nl = 0

⇒ 18nl + 30n^{2} + 15l^{2} + 25ln + 2nl = 0

⇒ 30n^{2} + 15l^{2} + 45nl = 0

⇒ 3l^{2} + 6n^{2} + 9nl = 0

⇒ l^{2} + 3nl + 2n^{2} = 0

⇒ \(\left(\frac{l}{n}\right)^2+3\left(\frac{l}{n}\right)\) + 2 = 0

⇒ \(\frac{l}{n}\) = – 1, – 2

\(\frac{l}{n}\) = – 1

⇒ l + 0m + n = 0

and 3l + m + 5n = 0

i.e. \(\frac{l}{-1}=\frac{m}{-2}=\frac{n}{1}\)

\(\frac{l}{n}\) = – 2

l + 0m + 2n = 0

3l + m + 5n = 0

i.e. \(\frac{l}{-2}=\frac{m}{1}=\frac{n}{1}\)

∴ D’ratios of both lines are < – 1, – 2, 1 > and < – 2, 1, 1 >

If θ be the acute angle between the lines.

Then cos θ = \(\frac{|(-1)(-2)-2(1)+1(1)|}{\sqrt{(-1)^2+(-2)^2+1^2} \sqrt{(-2)^2+1^2+1^2}}\)

= \(\frac{1}{\sqrt{6} \sqrt{6}}=\frac{1}{6}\)

θ = cos^{-1} (\(\frac{1}{6}\))

Question 20.

Find the point in which the join of A(- 9, 4, 5) and B (11, 0, – 1) is met by the perpendicular from the origin.

Solution:

Let N be the point in which join of A and B is met by the ⊥ drawn from origin and let N divides

AB in the ratio λ : 1

Question 21.

Find the coordinates of the foot of perpendicular drawn from the point A (1, 2, 1) on the line joining the points B (1, 4, 6) and C (5, 4, 4).

Solution:

Let N be the foot of J drawn from A (1, 2, 1) on line BC.

Let N divides line BC in the ratio λ : 1.

Question 22.

The figure shows a rectangular box of size a, b and c units along x, y and z-axes respectively. Prove that the angles between the four diagonals are given by cos^{-1} \(\left(\frac{ \pm a^2 \pm b^2 \pm c^2}{a^2+b^2+c^2}\right)\) where all signs cannot be + ve or – ve.

Solution:

Let O be the origin.

OX, OY and OZ be the three mutually ⊥ axes.

Here, OA, OB and OC are the coterminus edges of the rectangular parallelopiped s.t. OA = a ; OB = b and OC = c.

Then the coordinates of other vertices are ;

O (0, 0, 0); P (a, b, c) ; A (a, 0, 0) ; B (0, b, 0) ;

C (0, 0, c), L (a, 0, c) ; M (a, b, 0) and N (0, b, c).

Clearly OP, AN, BL and CM are the four diagonals of rectangular parallelopiped.

∴ D’ratios of OP are < a – 0, b – 0, c – 0 > i.e.< a, b, c >

D’ ratios of AN are < – a, b, c >

D’ ratios of BL are <a, – b, c >

∴ D’ ratios of diagonal CM are < a, b, – c>

Let a be the angle between OP and AN

Then cos α = \(\frac{a(-a)+b(b)+c(c)}{\sqrt{a^2+b^2+c^2} \sqrt{(-a)^2+b^2+c^2}}\)

= \(\frac{-a^2+b^2+c^2}{a^2+b^2+c^2}\)

∴ α = cos^{-1} \(\left(\frac{-a^2+b^2+c^2}{a^2+b^2+c^2}\right)\)

Let β be the angle between OP and BL

Then cos β = \(\frac{a(a)+b(-b)+c(c)}{\sqrt{a^2+b^2+c^2} \sqrt{a^2+(-b)^2+c^2}}\)

= \(\frac{a^2-b^2+c^2}{a^2+b^2+c^2}\)

∴ β = cos^{-1} \(\left(\frac{a^2-b^2+c^2}{a^2+b^2+c^2}\right)\)

Let γ be the angle between OP and CM

Then cos γ = \(\frac{a(a)+b(b)+c(-c)}{\sqrt{a^2+b^2+c^2} \sqrt{a^2+b^2+(-c)^2}}\)

= \(\frac{a^2+b^2-c^2}{a^2+b^2+c^2}\)

∴ γ = cos^{-1} \(\left(\frac{a^2+b^2-c^2}{a^2+b^2+c^2}\right)\)

Similarly the angles between the other pairs of diagonals can be find out.

similarly cos θ_{4} = \(\frac{-a^2-b^2+c^2}{a^2+b^2+c^2}\)

where θ_{4} is the angle between AL and BM

If θ_{5} is the between AL and CN

∴ cos θ_{5} = \(\frac{-a^2+b^2-c^2}{a^2+b^2+c^2}\)

If θ_{6} be the angle between BM and CN

∴ cos θ_{6} = \(\frac{a^2-b^2-c^2}{a^2+b^2+c^2}\)

Hence the angle between the four diagonals are given by cos^{-1} \(\left(\frac{ \pm a^2 \pm b^2 \pm c^2}{a^2+b^2+c^2}\right)\).

Question 22 (old).

Prove that the lines whose direction cosines are given by

ul^{2} + vm^{2} + wn^{2} = 0 and al + bm + cn = 0 are

(1) perpendicular if u (b^{2} + c^{2}) + v (c^{2} + a^{2}) + w (a^{2} + b^{2}) = 0

(2) parallel if \(\frac{a^2}{u}+\frac{b^2}{v}+\frac{c^2}{w}\) = 0

Solution:

The eqns. giving direction cosines of lines are

al + bm + cn = 0 …………….( 1)

and ul^{2} + vm^{2} + wn^{2} = 0 ……………(2)

From (1) ;

n = – \(\frac{(a l+b m)}{c}\)

∴ from (2);

we have ul^{2} + vm^{2} + w(\(\left(\frac{a l+b m}{c}\right)^2\)) = 0

⇒ c^{2} ul^{2} + c^{2} vm^{2} + w (al + bm)^{2} = 0

⇒ (a^{2} w + c^{2}u) l^{2} + (2ab lm w) + (c^{2} v + b^{2} w) m^{2} = 0

⇒ (a^{2} w + c^{2} u) (\(\frac{l}{m}\))^{2} + 2abw \(\frac{l}{m}\) + c^{2} v + b^{2} w = 0 …………….(3)

which is quadratic in \(\frac{l}{m}\) and giving two values of \(\frac{l}{m}\).

lf < l_{1}, m_{1}, n_{1} > and < l_{2}, m_{2}, n_{2} > be the direction cosines of two lines given by (1) and (2) then

\(\frac{l_1}{m_1}\) and \(\frac{l_2}{m_2}\) are the roots of eqn. (5).

∴ product of roots = \(\frac{l_1}{m_1} \cdot \frac{l_2}{m_2}\)

= \(\frac{c^2 v+b^2 w}{a^2 w+c^2 u}\)

⇒ \(\frac{l_1 l_2}{c^2 v+b^2 w}=\frac{m_1 m_2}{a^2 w+c^2 u}=\frac{n_1 n_2}{b^2 u+a^2 v}\) = k (say) ≠ 0 [by symmetry]

∴ l_{1}l = k (c^{2}v + b^{2}w) ;

m_{1}m_{2} = k (a^{2}w + c^{2}u)

n_{1}n_{2} = k (b^{2}u + a^{2}v)

(i) Now, the lines given by (1) and (2) whose direction cosines are < l_{1}, m_{1}, n_{1} > and < l_{2}, m_{2}, n_{2} > are perpendicular

if l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 0

if k [c^{2}v + b^{2}w + a^{2}w + c^{2}u + b^{2}2u + a^{2}v] = 0

if u (c^{2} + b^{2}) + v (c^{2} + a^{2}) + (a^{2} + b^{2}) = 0 [ k ≠ 0]

(ii) The lines given by (1) and (2) are parallel if their direction cosines are same.

So the roots (*) are equal.

∴ discriminant of eqn. (*) is equal to 0.

if 4a^{2}b^{2}w^{2} – 4 (a^{2}w + c^{2}u) (c^{2}v + b^{2}w) = 0

if 4a^{2}b^{2}w^{2} – 4a^{2}b^{2}w^{2} – 4a^{2}c^{2}vw – 4c^{4}uv – 4b^{2}c^{2}uw = 0

if a^{2}c^{2}vw + c^{4}uv + b^{2}c^{2}uw = 0

if a^{2}vw + c^{2}uv + b^{2}uw = 0

if \(\frac{a^2}{u}+\frac{b^2}{v}+\frac{c^2}{w}\) = 0

[On dividing throughout by uvw]