If x y z is on the plane π we have x 1 3 t 3 s y 1 t

If ( x, y, z ) is on the plane π , we have



x = − 1 + 3 t + 3 s
y = 1 − t − 2 s
z = 2 − 5 t
. The third equa-
tion gives t =
2 − z
5
. If we substitute this into the second equation we get
y = 1 −
2 − z
5
− 2 s which gives s = −
1
2
y +
3
10
+
z
10
. We substitute the values for
t and s into the first equation giving x = − 1 + 3
parenleftbigg
2 − z
5
parenrightbigg
+ 3
parenleftbigg
−
1
2
y +
3
10
+
z
10
parenrightbigg
.
Rewriting we get 10 x + 15 y + 3 z = 11 as an equation for π .
(An alternate approach would be to find n = v × w .)
(c) A direction vector for this line is a normal vector for the plane π , n = (10 , 15 , 3).
A parametric description of the line is (0 , 1 , 0) + t (10 , 15 , 3), t ∈ R .
To see where the line and the plane meet we must find a t such that (10 t, 1+15 t, 3 t )
satisfies the equation of the plane. For this we need 10(10 t ) + 15(1 + 15 t ) +
3(3 t ) = 11 = ⇒ 334 t = − 4 = ⇒ t =
− 2
167
. The point of intersection is
parenleftbigg
−
20
167
,
137
167
, −
6
167
parenrightbigg
.

MATB41H Solutions # 4 page 3
3. Let ( p, q ) be the point where the perpendicular
from ( x
1
, y
1
) meets the line ℓ . A normal for ℓ is
n = ( a, b ). Hence ( p − x, q − y ) = t ( a, b ), for some
t ∈ R . The required distance is bardbl t ( a, b ) bardbl . We
can also write ( p, q ) as ( p, q ) = ( x
1
+ t a, y
1
+ t b )
and as a p + b q = c (since ( p, q ) is a point on ℓ ).
5. The tangent plane to the graph of z = f ( x, y ) at the point ( a, b, f ( a, b )) is
z = f ( a, b ) +
∂f
∂x
( a, b ) ( x − a ) +
∂f
∂y
( a, b ) ( y − b ).