The lines {r}={i}+{j}-{k}+λ({3i}-{j}) and {r}={4i}-{k}+({2i}+{3k}) intersect at the point

Correct Answer is : (4,0,−1)(4,0,-1)(4,0,−1)

Correct Answer is : (4,0,−1)(4,0,-1)(4,0,−1)

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KEAM 2011 Math Paper

Question : 27 of 120

Marks: +1, -0

The lines

\vec{r}=\hat{i}+\hat{j}-\hat{k}+\lambda(\hat{3i}-\hat{j})

\vec{r}=\vec{4i}-\hat{k}+\mu(\hat{2i}+\hat{3k})

intersect at the point

$(0,0,0)$
$(0,0,1)$
$(0,-4,-1)$
$(4,0,-1)$
$(4,1,-1)$

Solution:

The given lines,

r=(i+j-k)+\lambda(3i-j)

......(i)

and

r=(4i-k)+\mu(2i+3k)

.....(ii)

From Eqs. (i) and (ii),

(i+j-k)+\lambda(3i-j)

=(4i-k)+\mu(2i+3k)

(1+3\lambda)i+(1-\lambda)j-k

=(4+2\mu)i+0j+(-1+3\mu)k

Equating the coefficient of

i,j

and

k

on both sides,

1+3\lambda=4+2\mu

3\lambda-2\mu=3

......(iii)

1-\lambda=0

\Rightarrow\lambda=1

-1+3\mu=-1

\Rightarrow\mu=0

On putting these values in Eq. (i),

r=(i+j-k)+(3i-j)

r=4i+0j-k

So, the intersection point is

(4,0,-1)

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