RD Sharma Class 12 Ex 29.14 Solutions Chapter 29 The Plane

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Here we provide RD Sharma Class 12 Ex 29.14 Solutions Chapter 29 The Plane for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 29.14 Solutions Chapter 29 The Plane book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter29
Exercise29.14
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 29.14 Solutions Chapter 29 The Plane

Question 1. Find the shortest distance between the lines \frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}  and \frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2} .

Solution: 

Let us consider

P_1=\frac{x-2}{-1}=\frac{y-5}{2}=\frac{z-0}{3}\\ P_2=\frac{x-0}{2}=\frac{y+1}{-1}=\frac{z-1}{2}

According to the equations line P1 passes through the point P(2, 5, 0)

And the equation of a plane containing line P2 is

a(x – 0) + b(y + 1) + c(z – 1) = 0          -(1)

Where 2a – b + 2c = 0

If it is parallel to line P1 then

-a + 2b + 3c = 0

So, 

\frac{a}{-7}=\frac{b}{-8}=\frac{c}{3}

Now, substitute the value of a, b, c in the eq(1) we get

a(x – 0) + b(y + 1) + c(z – 1) = 0

-7(x – 0) – 8(y + 1) + 3(z – 1) = 0

-7x – 8y – 8 + 3z – 3 = 0

7x + 8y – 3z + 11 = 0          -(2)

So, this is the equation of the plane that contain line P2 and parallel to line P1.

Hence, the shortest distance between P1 and P2 = Distance between point P(2, 5, 0) and plane (2)

=\left|\frac{14+40+11}{\sqrt{7^2+8^2+(-3)^2}}\right|=\frac{65}{\sqrt{122}}

Question 2. 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: 

Let us consider

P_1:\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}
P_2:\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Let us assume the equation of the plane containing P1 is a(x + 1) + b(y + 1) + c(z+1) = 0

Plane is parallel to P1 = 7a – 6b + c = 0          -(1)

Plane is parallel to P2 = a – 2b + c = 0          -(2)

On solving eq(1) and eq(2), we get,

\frac{a}{-6+2}=\frac{b}{1-7}=\frac{c}{-14+6}\\ \frac{a}{-4}=\frac{b}{-6}=\frac{c}{-8}

The equation of the plane is -4(x + 1) – 6(y + 1) – 8(z + 1) = 0

Final equation of plane is 4(x + 1) + 6(y + 1) + 8(z + 1) = 0 

Question 3. Find the shortest distance between the lines \frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}  and 3x – y – 2z + 4 = 0, 2x + y + z + 1 = 0.

Solution: 

The equation of a plane containing the line 3x – y – 2z + 4 = 0, 2x + y + z + 1 = 0 is 

x(2λ + 3) + y(λ – 1) + z(λ – 2) + λ + 4 = 0          -(1)

If it is parallel to the line \frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}  then,

2(2λ + 3) + 4(λ – 1) + (λ – 2) = 0

λ = 0

On putting λ = 0 in eq(1) we get,

3x – y – 2z + 4 = 0          -(2)

As this equation of the plane consist the second line and parallel to the first line.

It is clear that the line \frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}  passes through the point (1, 3, -2)

So, the shortest distance ‘D’ between the given lines is equal to the 

length of perpendicular from point (1, 3, -2) on the plane (2)

D = \left|\frac{3-3+4+4}{\sqrt{1+9+4}}\right|=\frac{8}{\sqrt{14}}

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