Question

If $L_{1}$ is a line through the point $5 \hat{ i }+8 \hat{ j }+11 \hat{ k }$ and parallel to the vector $2 \hat{ i }+3 \hat{ j }+4 \hat{ k }$ and $L_{2}$ is a line through the point $4 \hat{ i }+6 \hat{ j }+8 \hat{ k }$ and parallel to the vector $3 \hat{ i }+4 \hat{ j }+5 \hat{ k }$, then the point of intersection of $L_{1}$ and $L_{2}$ is

• A. $\hat{ i }+\hat{ j }+\hat{ k }$
• B. $\hat{ i }+2 \hat{ j }+3 \hat{ k }$
• C. $2 \hat{ i }+3 \hat{ j }+\hat{ k }$
• D. $\hat{ i }-2 \hat{ j }+2 \hat{ k }$

Answer 1

Answer: (B)

Line $L_{1}$ is passing through $5 \hat{ i }+8 \hat{ j }+11 \hat{ k }$ and parallel to the vector $2 \hat{ i }+3 \hat{ j }+4 \hat{ k }$.
$\therefore \, L_{1}= r =5 \hat{ i }+8 \hat{ j }+11 \hat{ k }+\lambda(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })$
line $L_{2}$ is passing through $4 \hat{ i }+6 \hat{ j }+8 \hat{ k }$ and parallel to the vector $3 \hat{ i }+4 \hat{ j }+5 \hat{ k }$
$\therefore \,L_{2}= r =4 \hat{ i }+6 \hat{ j }+8 \hat{ k }+\mu(3 \hat{ i }+4 \hat{ j }+5 \hat{ k })$
$L_{1}$ and $L_{2}$ are intersecting.
$\therefore \, 5 \hat{ i }+8 \hat{ j }+11 \hat{ k }+\lambda(2 \hat{ i }+3 \hat{ j }+4 \hat{ k })=4 \hat{ i }+6 \hat{ j }$
$+8 \hat{ k }+\mu(3 \hat{ i }+4 \hat{ j }+5 \hat{ k })$
$\Rightarrow \hat{ i }+2 \hat{ j }+3 \hat{ k }=(3 \mu-2 \lambda) \hat{ i }+(4 \mu-3 \lambda) \hat{ j }+(5 \mu-4 \lambda) \hat{ k }$
$\Rightarrow \, 3 \mu-2 \lambda=1\,\,\,\,\,\,\dots(i)$
$4 \mu-3 \lambda=2 \,\,\,\,\,\,\dots(ii)$
On solving Eqs. (i) and (ii), we get
$\mu=-1, \lambda=-2$
Put $ \lambda=-2$ in $L_{1}$
We get $ r =\hat{ i }+2 \hat{ j }+3 \hat{ k }$
$\therefore $ Point of intersection of $L_{1}$ and $L_{2}$ is $\hat{ i }+2 \hat{ j }+3 \hat{ k }$