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NCERT solution class 12 chapter 5 Three Dimensional Geometry exercise 5.2 mathematics part 2

EXERCISE 5.2


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Question 1:

Show that the three lines with direction cosines

 are mutually perpendicular.

Answer:

Two lines with direction cosines, l1m1n1 and l2m2n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0

(i) For the lines with direction cosines,  and , we obtain

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines,  and , we obtain

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines,  and , we obtain

Therefore, the lines are perpendicular.

Thus, all the lines are mutually perpendicular.

Question 2:

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

Answer:

Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6).

The direction ratios, a1b1c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4.

The direction ratios, a2b2c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.

AB and CD will be perpendicular to each other, if a1a2 + b1b2c1c2 = 0

a1a2 + b1b2c1c2 = 2 × 3 + 5 × 2 + (− 4) × 4

= 6 + 10 − 16

= 0

Therefore, AB and CD are perpendicular to each other.

Question 3:

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

Answer:

Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5).

The directions ratios, a1b1c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4.

The direction ratios, a2b2c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4.

AB will be parallel to CD, if 

Thus, AB is parallel to CD.

Question 4:

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector.

Answer:

It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is 

It is known that the line which passes through point A and parallel to is given by is a constant.

This is the required equation of the line.

Question 5:

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector  and is in the direction .

Answer:

It is given that the line passes through the point with position vector

It is known that a line through a point with position vector and parallel to is given by the equation, 

This is the required equation of the line in vector form.

Eliminating λ, we obtain the Cartesian form equation as

This is the required equation of the given line in Cartesian form.

Question 6:

Find the Cartesian equation of the line which passes through the point

(−2, 4, −5) and parallel to the line given by

Answer:

It is given that the line passes through the point (−2, 4, −5) and is parallel to 

The direction ratios of the line, are 3, 5, and 6.

The required line is parallel to 

Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0

It is known that the equation of the line through the point (x1y1z1) and with direction ratios, abc, is given by 

Therefore the equation of the required line is

Question 7:

The Cartesian equation of a line is . Write its vector form.

Answer:

The Cartesian equation of the line is

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

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

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

It is known that the line through position vector and in the direction of the vector is given by the equation, 

This is the required equation of the given line in vector form.

Question 8:

Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).

Answer:

The required line passes through the origin. Therefore, its position vector is given by,

The direction ratios of the line through origin and (5, −2, 3) are

(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3

The line is parallel to the vector given by the equation, 

The equation of the line in vector form through a point with position vector and parallel to is, 

The equation of the line through the point (x1y1z1) and direction ratios abc is given by, 

Therefore, the equation of the required line in the Cartesian form is

Page No 478:

Question 9:

Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

Answer:

Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its position vector is given by,

The direction ratios of PQ are given by,

(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11

The equation of the vector in the direction of PQ is

The equation of PQ in vector form is given by, 

The equation of PQ in Cartesian form is

 i.e.,

Question 10:

Find the angle between the following pairs of lines:

(i) 

(ii) and

Answer:

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by, 

The given lines are parallel to the vectors, and , respectively.

(ii) The given lines are parallel to the vectors, and , respectively.

Question 11:

Find the angle between the following pairs of lines:

(i) 

(ii) 

Answer:

  1. Let and  be the vectors parallel to the pair of lines, respectively.

and 

The angle, Q, between the given pair of lines is given by the relation,

(ii) Let  be the vectors parallel to the given pair of lines,  and , respectively.

If Q is the angle between the given pair of lines, then 

Question 12:

Find the values of p so the line and

are at right angles.

Answer:

The given equations can be written in the standard form as

 and 

The direction ratios of the lines are −3,, 2 and  respectively.

Two lines with direction ratios, a1b1c1 and a2b2c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0

Thus, the value of p is .

Question 13:

Show that the lines and are perpendicular to each other.

Answer:

The equations of the given lines areand 

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.

Two lines with direction ratios, a1b1c1 and a2b2c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0

∴ 7 × 1 + (−5) × 2 + 1 × 3

= 7 − 10 + 3

= 0

Therefore, the given lines are perpendicular to each other.

Question 14:

Find the shortest distance between the lines

Answer:

The equations of the given lines are

It is known that the shortest distance between the lines,  and , is given by,

d = b1→×b2→.a2→-a1→b1→×b2→

Comparing the given equations, we obtain

Substituting all the values in equation (1), we obtain

Therefore, the shortest distance between the two lines is units.

Question 15:

Find the shortest distance between the lines  and 

Answer:

The given lines are  and 

It is known that the shortest distance between the two lines, is given by,

Comparing the given equations, we obtain

Substituting all the values in equation (1), we obtain

Since distance is always non-negative, the distance between the given lines is units.

Question 16:

Find the shortest distance between the lines whose vector equations are

Answer:

The given lines are and 

It is known that the shortest distance between the lines,  and , is given by,

Comparing the given equations with  and , we obtain 

Substituting all the values in equation (1), we obtain

Therefore, the shortest distance between the two given lines is units.

Question 17:

Find the shortest distance between the lines whose vector equations are

Answer:

The given lines are

r→=(s+1)i^+(2s-1)j^-(2s+1)k^⇒r→=(i^-j^-k^)+s(i^+2j^-2k^)             …(2)It is known that the shortest distance between the lines,  and , is given by,

For the given equations,

Substituting all the values in equation (3), we obtain

Therefore, the shortest distance between the lines isunits.


 

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