Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2

Question 1.

Show that the three 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 perpendicular.

Solution:

The line with direction cosines 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.

(i) Line with direction cosines \(\frac{12}{13}\), \(\frac{- 3}{13}\), \(\frac{- 4}{13}\) and \(\frac{4}{13}\), \(\frac{12}{13}\), \(\frac{3}{13}\)

are perpendicular, if

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

or, if \(\frac{48-36-12}{169}\) = 0, which is true.

⇒ These lines are perpendicular.

(ii) The lines with direction cosines \(\frac{4}{13}\), \(\frac{12}{13}\), \(\frac{3}{13}\) and \(\frac{3}{13}\), \(\frac{- 4}{13}\), \(\frac{12}{13}\)

are perpendicular, if

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

\(\frac{12-48+36}{169}\) = 0, which is true.

⇒ These lines are perpendicular.

(iii) the lines with direction cosines \(\frac{3}{13}\), \(\frac{- 4}{13}\), \(\frac{12}{13}\)

and \(\frac{12}{13}\), \(\frac{- 3}{13}\), \(\frac{- 4}{13}\) are perpendicular, if

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

= \(\frac{36+12-48}{169}\) = 0, which is true.

∴ These lines are perpendicular.

⇒ Given lines are mutually perpendicular.

Question 2.

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).

Solution:

Let A and B be the points (1, – 1, 2) and (3, 4, – 2) respectively.

Direction ratios of AB are

3 – 1, 4 – (- 1), – 2 – 2 or 2, 5, – 4.

Let C and D be the points (0, 3, 2) and (3, 5, 6) respectively.

Direction ratios of CD are

3 – 0, 5 – 3, 6 – 2 or 3, 2, 4.

AB is perpendicular to CD, if a_{1}a_{2} + b_{1}b_{2} + C_{1}C_{2} = 0.

i.e., 2 × 3 + 5 × 2 + (- 4) × 4

= 6 + 10 – 16 = 0, which is true.

⇒ AB ⊥ CD.

Question 3.

Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (- 1, – 2, 1) and (1, 2, 5).

Solution:

Let A and B be the points (4, 7, 8) and (2, 3, 4) respectively.

Direction ratios of AB are 2 – 4, 3 – 7, 4 – 8 or – 2, – 4, – 4 or 1, 2, 2.

Let C and D be the points (- 1, – 2, 1) and (1, 2, 5) respectively.

∴ Direction ratios of CD are

1 – (- 1), 2 – (- 2), 5 – 1 or 2, 4, 4.

AB is parallel to CD if \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\).

Here, \(\frac{1}{2}\) = \(\frac{2}{4}\) = \(\frac{2}{4}\), which is true.

Hence, AB || CD.

Question 4.

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3\(\hat {i} \) + 2\(\hat {j} \) – 2\(\hat {k} \).

Solution:

Equation of the line passing through the point a and parallel to vector \(\vec{b}\) is

\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)

Here, \(\vec{a}\) = \(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \) and \(\vec{b}\) = 3\(\hat {i} \) + 2\(\hat {j} \) – 2\(\hat {k} \).

∴ Equation of the required line is

\(\vec{r}\) = (\(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \)) + λ(3\(\hat {i} \) + 2\(\hat {j} \) – 2\(\hat {k} \)), λ ∈ R.

Question 5.

Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2\(\hat {i} \) – \(\hat {j} \) + 4\(\hat {k} \)

and is in the direction \(\hat {i} \) + 2\(\hat {j} \) – \(\hat {k} \).

Solution:

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

\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)

Here, \(\vec{a}\) = 2\(\hat {i} \) – \(\hat {j} \) + 4\(\hat {k} \) and \(\vec{b}\) = \(\hat {i} \) + 2\(\hat {j} \) – \(\hat {k} \).

So, the vector equation of the required line is

\(\vec{r}\) = (2\(\hat {i} \) – \(\hat {j} \) + 4\(\hat {k} \)) + λ(\(\hat {i} \) + 2\(\hat {j} \) – \(\hat {k} \)) ……………. (1),

where λ is parameter.

In Cartesiant Form :

Putting \(\vec{r}\) = x\(\hat {i} \) + y\(\hat {j} \) + z\(\hat {k} \) in (1), we get

x\(\hat {i} \) + y\(\hat {j} \) + z\(\hat {k} \) = (2\(\hat {i} \) – \(\hat {j} \) + 4\(\hat {k} \)) + λ(\(\hat {i} \) + 2\(\hat {j} \) – \(\hat {k} \))

⇒ x\(\hat {i} \) + y\(\hat {j} \) + z\(\hat {k} \) = (2 + λ)\(\hat {i} \) + (- 1 + 2λ)\(\hat {j} \) + (4 – λ)\(\hat {k} \)

Equating coefficients of \(\hat {i} \), \(\hat {j} \) and \(\hat {k} \), we get

x = 2 + λ, y = – 1 + 2λ, z = 4 – λ

⇒ x – 2 = λ, \(\frac{y+1}{2}\) = λ, \(\frac{z-4}{-1}\) = λ

Eliminating λ, we have:

\(\frac{x-2}{1}\) = \(\frac{y+1}{2}\) = \(\frac{z-4}{-1}\).

Alternative Method :

The equation (1) represents a line passing through a point (2, – 1, 4) and has direction ratios 1, 2, – 1.

∴ The cartisian form of its equations is

\(\frac{x-2}{1}\) = \(\frac{y+1}{2}\) = \(\frac{z-4}{-1}\).

Question 6.

Find the cartesian equation of the line which passes through the point (- 2, 4, – 5) and parallel to the line given by

\(\frac{x+3}{3}\) = \(\frac{y-4}{5}\) = \(\frac{z+8}{6}\).

Solution:

The Cartesian equation of the line passing through the point (- 2, 4, – 5) and parallel to the line

\(\frac{x+3}{3}\) = \(\frac{y-4}{5}\) = \(\frac{z+8}{6}\) is \(\frac{x+2}{3}\) = \(\frac{y-4}{5}\) = \(\frac{z+5}{6}\).

Question 7.

The cartesian equation of a line \(\frac{x-5}{3}\) = \(\frac{y+4}{7}\) = \(\frac{z-6}{2}\). Write its vector form.

Solution:

The Cartesian equation of the line is

\(\frac{x-5}{3}\) = \(\frac{y+4}{7}\) = \(\frac{z-6}{2}\) ……………. (1)

Clearly (1) passes through the point A(5, – 4, 6) and has 3, 7, 2 as its direction ratios

⇒ Line (1) passes through the point A with position vector \(\vec{a}\) = 5\(\hat {i} \) – 4\(\hat {j} \) + 6\(\hat {k} \)

and is in the direction of \(\vec{b}\) = 3\(\hat {i} \) + 7\(\hat {j} \) + 2\(\hat {k} \).

So, the vector equation of the required line is

\(\vec{r}\) = (5\(\hat {i} \) – 4\(\hat {j} \) + 6\(\hat {k} \)) + λ(3\(\hat {i} \) + 7\(\hat {j} \) + 2\(\hat {k} \)).

Question 8.

Find the vector and cartesian equations of the line that passes through the origin and the point (5, – 2, 3).

Solution:

The line passes through the origin (0, 0, 0)

∴ \(\vec{a}\) = \(\vec{0}\)

Direction ratios of the line passing through the points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2})

are x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1}.

∴ Direction ratios of the line joining the points A(0, 0, 0) and B(5, – 2, 3) are

∴ 5 – 0, – 2 – 0, 3 – 0 or 5, – 2, 3.

∴ \(\vec{b}\) = 5\(\hat {i} \) – 2\(\hat {j} \) + 3\(\hat {k} \)

(i) Equation of line AB in vector form

\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)

= \(\vec{0}\) + λ(5\(\hat {i} \) – 2\(\hat {j} \) + 3\(\hat {k} \)).

(ii) Equation of the line AB in cartesian form is

\(\frac{x-x_{1}}{a}\) = \(\frac{y-y_{1}}{b}\) = \(\frac{z-z_{1}}{c}\),

where a, b, c are the direction ratios of the line,

which passes through (x_{1}, y_{1}, z_{1}).

Directon ratios are 5, – 2, 3 and the line passes through (0, 0, 0).

∴ Required equation of AB

\(\frac{x-0}{5}\) = \(\frac{y-0}{- 2}\) = \(\frac{z-0}{3}\)

or \(\frac{x}{5}\) = \(\frac{y}{- 2}\) = \(\frac{z}{3}\).

Question 9.

Find the vector and cartesian equations of the line that passes through the points P(3, – 2, – 5) and Q(3, – 2, 6).

Solution:

The line PQ passes through the point P(3, – 2, – 5).

∴ \(\vec{a}\) = 3\(\hat {i} \) – 2\(\hat {j} \) – 5\(\hat {k} \)

Direction ratios of PQ, where P and Q are (3, – 2, – 5) and (3, – 2, 6), are

x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1}

i.e., 3 – 3, – 2 – (- 2), 6 – (- 5) or 0, 0, 11

∴ \(\vec{b}\) = 11\(\hat {k} \)

(i) Equation of line PQ in vector form is

\(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)

= (3\(\hat {i} \) – 2\(\hat {j} \) – 5\(\hat {k} \)) + λ 11\(\hat {k} \)

or \(\vec{r}\) = (3\(\hat {i} \) – 2\(\hat {j} \) – 5\(\hat {k} \)) + λ 11\(\hat {k} \)

(ii) Equation of the line PQ in cartesian form is

\(\frac{x-x_{1}}{a}\) = \(\frac{y-y_{1}}{b}\) = \(\frac{z-z_{1}}{c}\),

P(x_{1}, y_{1}, z_{1}) is (3, – 2, – 5) and directon ratios a, b, c are 0, 0, 11.

∴ Equation of PQ is

\(\frac{x-3}{0}\) = \(\frac{y+2}{0}\) = \(\frac{z+5}{11}\).

Question 10.

Find the angle between the following pairs of lines:

(i) \(\vec{r}\) = 2\(\hat {i} \) – 5\(\hat {j} \) + \(\hat {k} \) + λ(3\(\hat {i} \) + 2\(\hat {j} \) + 6\(\hat {k} \)) and

\(\vec{r}\) = 7\(\hat {i} \) – 6\(\hat {k} \) + λ(\(\hat {i} \) + 2\(\hat {j} \) + 2\(\hat {k} \))

(ii) \(\vec{r}\) = 3\(\hat {i} \) + \(\hat {j} \) – 2\(\hat {k} \) + λ(\(\hat {i} \) – \(\hat {j} \) – 2\(\hat {k} \)) and

\(\vec{r}\) = 2\(\hat {i} \) – \(\hat {j} \) – 56\(\hat {k} \) + µ(3\(\hat {i} \) – 5\(\hat {j} \) – 4\(\hat {k} \))

Solution:

(i) Let θ be the angle between the given lines. The given lines are parallel to the vectors

\(\vec{b}_{1}\) = 3\(\hat {i} \) + 2\(\hat {j} \) + 6\(\hat {k} \) and

\(\vec{b}_{2}\) = \(\hat {i} \) + 2\(\hat {j} \) + 2\(\hat {k} \) respectively.

∴ The angle θ between them is given by

(ii) Let θ be the angle between the given lines. The given lines are parallel to the vectors

\(\vec{b}_{1}\) = \(\hat {i} \) – \(\hat {j} \) – 2\(\hat {k} \) and

\(\vec{b}_{2}\) = 3\(\hat {i} \) – 5\(\hat {j} \) – 4\(\hat {k} \) respectively.

∴ The angle θ between them is given by

Question 11.

Find the angle between the following pair of lines:

(i) \(\frac{x-2}{2}\) = \(\frac{y-1}{5}\) = \(\frac{z+3}{-3}\) and \(\frac{x+2}{-1}\) = \(\frac{y-4}{8}\) = \(\frac{z-5}{4}\).

(ii) \(\frac{x}{2}\) = \(\frac{y}{2}\) = \(\frac{z}{1}\) and \(\frac{x-5}{4}\) = \(\frac{y-2}{1}\) = \(\frac{z-3}{8}\).

Solution:

(i) Let \(\vec{b}_{1}\) and \(\vec{b}_{2}\) be the vectors parallel to

\(\frac{x-2}{2}\) = \(\frac{y-1}{5}\) = \(\frac{z+3}{-3}\) and \(\frac{x+2}{-1}\) = \(\frac{y-4}{8}\) = \(\frac{z-5}{4}\) respectively.

∴ \(\vec{b}_{1}\) = 2\(\hat {i} \) + 5\(\hat {j} \) – 3\(\hat {k} \)

and \(\vec{b}_{2}\) = – \(\hat {i} \) + 8\(\hat {j} \) + 4\(\hat {k} \)

(ii) The given equations are

\(\frac{x}{2}\) = \(\frac{y}{2}\) = \(\frac{z}{1}\) ……………… (1)

and \(\frac{x-5}{4}\) = \(\frac{y-2}{1}\) = \(\frac{z-3}{8}\) ……………….. (2)

Let \(\vec{b}_{1}\) and \(\vec{b}_{2}\) be the vectors parallel to (1) and (2) respectively.

Then, \(\vec{b}_{1}\) = 2\(\hat {i} \) + 2\(\hat {j} \) + \(\hat {k} \)

and \(\vec{b}_{2}\) = 4\(\hat {i} \) + \(\hat {j} \) + 8\(\hat {k} \).

If θ is the angle between the given lines, then

Question 12.

Find the value of p so that the lines

\(\frac{1-x}{3}\) = \(\frac{7y-14}{2p}\) = \(\frac{z-3}{2}\) and \(\frac{7-7x}{3p}\) = \(\frac{y-5}{1}\) = \(\frac{6-z}{5}\) are at right angles.

Solution:

The given equations are not in the standard form. The equations of the given lines can be respectively written as,

The direction ratios of the given lines are – 3, \(\frac{2p}{7}\), 2 and \(\frac{-3p}{7}\), 1, – 5.

The line are perpendicular to each other.

Question 13.

Show that the lines \(\frac{x-5}{7}\) = \(\frac{y+2}{-5}\) = \(\frac{z}{1}\)

and \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) are perpendicular to each other.

Solution:

The given lines are

\(\frac{x-5}{7}\) = \(\frac{y+2}{-5}\) = \(\frac{z}{1}\) …………….. (1)

and \(\frac{x}{1}\) = \(\frac{y}{2}\) = \(\frac{z}{3}\) …………….. (2)

Let \(\vec{b}_{1}\) = 7\(\hat {i} \) – 5\(\hat {j} \) + \(\hat {k} \)

and \(\vec{b}_{2}\) = \(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \).

If θ is the angle between the given lines, then

Hence, the given lines are perpendicular to each other.

Question 14.

Find the shortest distance between the lines

\(\vec{r}\) = (\(\hat {i} \) + 2\(\hat {j} \) + \(\hat {k} \)) + λ(\(\hat {i} \) – \(\hat {j} \) + \(\hat {k} \)) …………… (1)

and \(\vec{r}\) = (2\(\hat {i} \) – \(\hat {j} \) – \(\hat {k} \)) + µ(2\(\hat {i} \) + \(\hat {j} \) + 2\(\hat {k} \)) …………… (2)

The shortest distance between the lines

Comparing the equations (1) and (2) with \(\vec{r}\) = \(\vec{a}_{1}\) + λ\(\vec{b}_{1}\) and \(\vec{r}\) = \(\vec{a}_{2}\) + µ\(\vec{b}_{2}\) respectively, we have:

Question 15.

Find the shortest distance between the lines

\(\frac{x+1}{7}\) = \(\frac{y+1}{-6}\) = \(\frac{z+1}{1}\) and \(\frac{x-3}{1}\) = \(\frac{y-5}{-2}\) = \(\frac{z-7}{1}\).

Solution:

Shortest distance between the lines

Question 16.

Find the shortest distance the lines whose vector equations are

\(\vec{r}\) = (\(\hat {i} \) + 2\(\hat {j} \) + 3\(\hat {k} \)) + λ(\(\hat {i} \) – 3\(\hat {j} \) + 2\(\hat {k} \)) and …………… (1)

\(\vec{r}\) = (4\(\hat {i} \) + 5\(\hat {j} \) + 6\(\hat {k} \)) +µ(2\(\hat {i} \) + 3\(\hat {j} \) + \(\hat {k} \)) …………… (2)

We know that the shortest distance between the lines are

Comparing the equations (1) and (2) with \(\vec{r}\) = \(\vec{a}_{1}\) + λ\(\vec{b}_{1}\) and

\(\vec{r}\) = \(\vec{a}_{2}\) + µ\(\vec{b}_{2}\) respectively, we have:

Question 17.

Find the shortest distance between the lines whose vector equations are

\(\vec{r}\) = (1 – t)\(\hat {i} \) + (t – 2)\(\hat {j} \) + (3 – 2)\(\hat {k} \) ………………. (1) and

\(\vec{r}\) = (\(\hat {i} \) – \(\hat {j} \) – \(\hat {k} \)) + s(\(\hat {i} \) + 2\(\hat {j} \) – 2\(\hat {k} \)) …………… (2)

Solution:

The given equations can be written as

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

and \(\vec{r}\) = (\(\hat {i} \) – \(\hat {j} \) – \(\hat {k} \)) + s(\(\hat {i} \) + 2\(\hat {j} \) – 2\(\hat {k} \)) ……………… (2)

Comparing these equations with