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BITSAT 2017 Paper

Section: Mathematics

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Question : 125 of 150

Marks: +1, -0

If the line through the points

A(k,1,-1)

B(2k,0,2)

is perpendicular to the line through the points B and

C(2+2k,k,1)

, then what is the value of k?

–1
1
–3
3

Solution:

Given points are

A(k,1,-1),B(2k,0,2)

and

C(2+2k,k,1)

Let

r_1=

length of line

AB=\sqrt{(2k-k)^{2}+(0-1)^{2}+(2+1)^{2}}=\sqrt{k^{2}+10}

and

r_2=

length of line

BC=\sqrt{(2)^{2}+k^{2}+(-1)^{2}}

=\sqrt{k^{2}+5}

Now, let

l_1,m_1,n_1

be direction-cosines of line AB and

l_2,m_2,n_2

be the direction cosines of BC.

Since AB is perpendicular to BC

\therefore l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0

Now

l_{1}=\;\frac{k}{\sqrt{k^{2}+10}},\;m_{1}=\;\frac{-1}{\sqrt{k^{2}+10}},\;n_{1}=\;\frac{3}{\sqrt{k^{2}+10}}

and

l_{2}=\;\frac{2}{\sqrt{k^{2}+5}},\;m_{2}=\;\frac{k}{\sqrt{k^{2}+5}},\;n_{2}=\;\frac{-1}{\sqrt{k^{2}+5}}

So,

l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0

\Rightarrow\;\frac{2k}{\sqrt{k^{2}+10\sqrt{k^{2}+5}}}-\;\frac{k}{\sqrt{k^{2}+10\sqrt{k^{2}+5}}}-\;\frac{3}{\sqrt{k^{2}+10}\sqrt{k^{2}+5}}=0

\Rightarrow 2k-k-3=0

\Rightarrow k=3

For

k=3

, AB is perpendicular to BC.

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