The point of intersection of the lines l_1: r | vedclass

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(C) Let the point of intersection be $P$. Any point on line $l_1$ is given by $(1+t, -6+2t, 2+t)$.Any point on line $l_2$ is given by $(2u, 4+u, 1+2u)

Vedclass · Accepted Answer

$(8,8,9)$

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(C) Let the point of intersection be $P$. Any point on line $l_1$ is given by $(1+t, -6+2t, 2+t)$.Any point on line $l_2$ is given by $(2u, 4+u, 1+2u)$.For the lines to intersect, these points must be equal:$1+t = 2u$  $(i)$$-6+2t = 4+u$  (ii)$2+t = 1+2u$  (iii)From $(i)$, $t = 2u-1$. Substituting this into (ii):$-6+2(2u-1) = 4+u$$-6+4u-2 = 4+u$$3u = 12 \Rightarrow u = 4$.Substituting $u=4$ into $t=2u-1$, we get $t=2(4)-1 = 7$.Check with (iii): $2+7 = 9$ and $1+2(4) = 9$. The values are consistent.The point of intersection is $(1+7, -6+2(7), 2+7) = (8, 8, 9)$.

The point of intersection of the lines $l_1: r(t)=(i-6j+2k)+t(i+2j+k)$ and $l_2: R(u)=(4j+k)+u(2i+j+2k)$ is

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