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# If lines (x−1)/2=(y+1)/3=(z−1)/4 and  (x−3)/1=(y−k)/2=z/1 intersect, then find the value of k and hence find the equation of the plane containing these lines. - CBSE (Science) Class 12 - Mathematics

ConceptThree - Dimensional Geometry Examples and Solutions

#### Question

If lines (x−1)/2=(y+1)/3=(z−1)/4 and  (x−3)/1=(y−k)/2=z/1 intersect, then find the value of k and hence find the equation of the plane containing these lines.

#### Solution

he coordinates of any point on the first line are given by

(x−1)/2=(y+1)/3=(z−1)/4=lambda

i.e

x=2λ+1y=3λ1z=4λ+1

Thus, the coordinates of any point on this line are (2λ+1, 3λ1, 4λ+1).

The coordinates of any point on the second line are given by

(x−3)/1=(y−k)/2=z/1=μ

i.e.

x=μ+3

y=2μ+k

z=μ

Thus, the coordinates of any point on this line are (μ+3, 2μ+k, μ).

If these two lines intersect each other, then

2λ+1=μ+3, 3λ1=2μ+k, 4λ+1=μ

2λμ=2, 3λ2μ=k+1, 4λμ=1

Solving 2λμ=2 and 4λμ=1, we get

λ=3/2 and μ=5

By substituting the values λ=3/2 and μ=5 in 3λ2μ=k+1, we get

k=92

Also, we have

vecb_1=2hati+3hatj+4hatk and vecb_2=hati+2hatj+hatk

Now, the required plane contains both the given lines.

So, it passes through a point
veca (1, 1, 1) and perpendicular vector vecn , given by

vecN=vecb_1xxvecb_2

therefore vecN=[[hati,hatj,hatk],[2,3,4],[1,2,1]]

=>vecN=-5hati+2hatj+hatk

Therefore, the equation of plane passing through veca and perpendicular to vecN is given by

(vecr-veca).vecN=0

=>[vecr-(hati-hatj+hatk)].(-5hati+2hatj+hatk)=0

=>vecr.(-5hati+2hatj+hatk)=(hati-hatj+hatk).(-5hati+2hatj+hatk)

=>vecr.(-5hati+2hatj+hatk)=-6

or
5x − 2yz − 6 = 0

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Solution If lines (x−1)/2=(y+1)/3=(z−1)/4 and  (x−3)/1=(y−k)/2=z/1 intersect, then find the value of k and hence find the equation of the plane containing these lines. Concept: Three - Dimensional Geometry Examples and Solutions.
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