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

Let the line L be written in vector form as

$[a_0,b_0,c_0]+[a_1,b_1,c_1]t.$

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:

$\mathbf{(}\mathbf{𝟏}\mathbf{)}[2,1,-1]\cdot [a_1,b_1,c_1]=2a_1+b_1-c_1=0$

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:

$\mathbf{(}\mathbf{𝟐}\mathbf{)}[-1,-2,3]\cdot [a_1,b_1,c_1]=-a_1-2b_1+3c_1=0$

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

$-a_1-2b_1+3(2a_1+b_1)=0\to b_1=-5a_1.$

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.