The Line L1 passes through the point P(2,4,8) and Q(4,5,4).The line L2 is perpendicular to L1,and parallel to (3p 2p 4),where p∈ ℤ.
Find the value of p.Then,given that L2 passes through R(10,6,-40),write down a vector equation for L2.Also,the lines L1 and L2 intersect at the point A.What is the x-coordinate of A?
1 Answer | Add Yours
You need to write the vectors `barv_1 ` and `barv_2` of the lines L1 and L2 such that:
`bar v_1 = (4-2,5-4,4-8) =gt bar v_1 =(2,1,-4)`
`` `bar v_2 = (x,y,z)`
You need to remember that the vector `bar v_2` is parallel to (3p, 2p, 4) if:
`[[bar i, bar j, bar k],[x , y, z], [3p, 2p, 4]] = 0` => `4y bar i+ 2pxbar k + 3pz bar j - 3py bar k - 2pz bar i - 4x bar j`
Grouping the terms containing `bar i, bar j, bar k` yields:
`bar i(4y - 2pz) + bar j (3pz - 4x) + bar k(2px - 3py) = 0`
Equating the coefficients of `bar i, bar j, bar k` :
`4y - 2pz = 0 =gt 4y = 2pz =gt 2y = pz =gt y = pz/2`
`3pz - 4x = 0 =gt 3pz = 4x =gt x = 3pz/4`
`2px - 3py = 0 =gt 2px = 3py =gt 2x = 3y =gt x = 3y/2`
`3pz/4 = 3y/2 =gt pz/2 = y`
The vectors `bar v_1` and`barv_2` `v_` are orthogonal if dot product is zero:
Hence `bar v_2 = (3pz/4,pz/2,z)`
`bar v_1*bar v_2 = 0`
`bar v_1*bar v_2 = 2*(3pz)/4+ pz/2 - 4z = 0`
4z = 0 => z = 0
Since z= 0, p may take any value for`L_1` and`L_2` to be orthogonal.
We’ve answered 318,988 questions. We can answer yours, too.Ask a question