The position vector of the points `Pa n dQ` are `5 hat i+7 hat j-2 hat k` and `-3 hat i+3 hat j+6 hat k` , respectively. Vector ` vec A=2 hat i- hat j+ hat k` passes through point `P` and vector ` vec B=3 hat i+2 hat j+4 hat k` passes through point `Q` . A third vector `2 hat i+7 hat j-5 hat k` intersects vectors `Aa n dBdot` Find the position vectors of points of intersection.

Answer

`2hati +8hatj-3hatk, hatj +2hatk`

Explanation

Let vector `2hati+7hatj-5hatk` intersect vectors `vecA and vecB` at points L and M, respectively, which have to be determined. Take them to be `(x_1, y_1, z_1) and (x_2, y_2, z_2)` respectively.

PL is collinear with vector `vecA`. Therefore,
`" "vec(PL) = lamda vecA`
Comparing the coefficient of `hati, hatj and hatk`, we get
`" "(x_1-5)/(3) = (y_1-7)/(-1)= (z_1+2)/(1) = lamda ` (say)
L is `3lamda +5, -lamda+7, lamda-2`
Similarly, `vec(QM)= mu vecB`. Therefore,
`" " (x_3+3)/(-3)= (y_2-3)/(2) = (z_2-6)/(4)= mu `(say)
Therefore, M is `-3mu-3, 2mu+3, 4mu+6`
Again LM is collinear with vector `2hati+7hatj-5hatk`.
Therefore,
`(x_2-x_1)/(2)= (y_2-y_1)/(7) = (z_2-z_1)/(-5)= v` (say)
`(-3mu-3lamda -8)/(2) = (2mu+lamda-4)/( 7) = (4mu-lamda +8)/(-5)=v`
`3mu+3lamda +2v=-8`
`2mu+lamda -7v=4`
`4mu-lamda +5v=-8`
Solving, we get
`lamda =mu = v=-1`
Therefore, point L is `(2, 8, -3) or 2hati+8hatj-3hatk` and M is `(0, 1, 2) or hatj+2hatk`