If the line through the points A (k,1,-1) and B(2k,0,2) is perpendicular to the line through the points B and `C(2+2,k,1)`, then what is the value of k?

Options

A. `-1`
B. `1`
C. `-3`
D. `3`

Answer

D

Explanation

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
`:. l_1l_2+m_1m_2+n_1n_2=0`
Now, `l_1=k/sqrt(k^2+10),m_1=(-1)/sqrt(k^2+10),n_1=3/sqrt(k_2+10)`
`and l_2 =2/sqrt(k^2+5),m_2=k/sqrt(k^2+5),n_2=1/sqrt(k^2+5)`
So, `l_1l_2+m_1m_2+n_1n_2=0`
`rArr (2k)/(sqrt(k^2+10)sqrt(k^2+5))-k/(sqrt(k^2+10)sqrt(k^2+5))-3/(sqrt(k^2+10)sqrt(k^2+5))=0`
`rArr 2k-k-3=0`
`rArr k=3`
For k = 3, AB is perpendicular to BC.