Science:Math Exam Resources/Courses/MATH200/April 2012/Question 01 (a)/Solution 1

Let the line L be written in vector form as

$\displaystyle [a_{0},b_{0},c_{0}]+[a_{1},b_{1},c_{1}]t.$

If L is parallel to the plane 2x + y - z = 5, then L must be perpendicular to the normal vector of said plane, n = [2,1,-1]. i.e. The dot product between n and the direction of L must be equal to zero:

$[2,1,-1]\cdot [a_{1},b_{1},c_{1}]=0.$

Evaluating the dot product gives:

$\mathbf {(1)} \quad [2,1,-1]\cdot [a_{1},b_{1},c_{1}]=2a_{1}+b_{1}-c_{1}=0$

Remember that we also need the vector is perpendicular to the line,$[3-t,1-2t,3t]=[3,1,0]+[-1,-2,3]t.$ Hence, the following equation must also be satisfied:

$\displaystyle \mathbf {(2)} \quad [-1,-2,3]\cdot [a_{1},b_{1},c_{1}]=-a_{1}-2b_{1}+3c_{1}=0$

So we have two equations and three unknowns. This means there are infinitely many vectors that will satisfy the given conditions. We only need to find one. (Note: Clearly $a_{1}=b_{1}=c_{1}=0$ is a solution to the above equations (1), (2), but it is the trivial solution and is perpendicular to every vector, including the direction of the line L.)

Solving (1) for $c_{1}$ gives $c_{1}=2a_{1}+b_{1}$. Subbing this into (2) gives

$\displaystyle -a_{1}-2b_{1}+3(2a_{1}+b_{1})=0\quad \rightarrow \quad b_{1}=-5a_{1}.$

From this we get that $c_{1}=2a_{1}-5a_{1}=-3a_{1}$.

Thus, any vector of the form k[1,-5,-3] where k is a constant (not equal to zero) will be parallel to L. For example, the vectors [1,-5,-3] and [-2,10,6] are acceptable solutions.