RD Sharma Solutions Class 12 Straight Line in Space Exercise 28.5

RD Sharma Solutions for Class 12 Maths Chapter 28 Straight Line in Space Exercise 28.5, is provided here for students to prepare well for their board exams. Tutors have designed the solutions using a unique and simple method through which any student can grasp the topics with ease. The RD Sharma Solutions to this chapter are readily available in PDF format, which can be downloaded by the students easily from the links provided below, and they can start practising offline to score good marks in their board exams.

RD Sharma Solution for Class 12 Maths Chapter 28 Exercise 5

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Q1.i

Solution:

Let us solve for b, we get

……… (3)

By solving (2) and (3), we get

Now let us substitute the above-obtained values in equation (1), to get the shortest distance between given lines,

Hence, the shortest distance is 3√30 units.

ii.

Solution:

Given:

The equation of the lines is,

We know that,

Shortest distance between lines
is given by

…….. (1)

Where,

So now,

……. (2)

Let us solve for b, we get

…… (3)

By solving (2) and (3), we get

= -512

Now let us substitute the above-obtained values in equation (1), to get the shortest distance between given lines,

Hence, the shortest distance is 512/√3968 units.

iii.

Solution:

Given:

The equation of the lines is,

We know that,

Shortest distance between lines
is given by

…….. (1)

Where,

So now,

…….. (2)

Let us solve for b, we get

……… (3)

By solving (2) and (3), we get

= 1

Now let us substitute the above-obtained values in equation (1), to get the shortest distance between given lines,

Hence, the shortest distance is 1/√6 units.

iv.

Solution:

Given:

The vector equations are

The above equations can be written as

By using the formula,

Hence,

The shortest distance is 9/3√2 = 3/√2 units.

v.

Solution:

Given:

The equation of lines is,

We know that,

Shortest distance between lines
is given by

…….. (1)

Where,

So now,

……. (2)

Let us solve for b, we get

By solving (2) and (3), we get

= 15

Now let us substitute the above-obtained values in equation (1), to get the shortest distance between given lines,

Hence, the shortest distance is 5/√2 units.