Find a vector-valued function whose graph is the plane $x + y + z = 4 \text {. }$
A) $\mathbf { r } ( u , v ) = u \mathbf { i } - v \mathbf { j } - ( 4 - u - v ) \mathbf { k }$
B) $\mathbf { r } ( u , v ) = u \mathbf { i } + v \mathbf { j } + ( 4 - u - v ) \mathbf { k }$
C) $\mathbf { r } ( u , v ) = u \mathbf { i } + v \mathbf { j } + ( 4 + u + v ) \mathbf { k }$
D) $\mathbf { r } ( u , v ) = u \mathbf { i } + v \mathbf { j } - ( 4 + u + v ) \mathbf { k }$
E) $\mathbf { r } ( u , v ) = - u \mathbf { i } - v \mathbf { j } + ( 4 - u - v ) \mathbf { k }$

Question

Find a vector-valued function whose graph is the plane $x + y + z = 4 \text {. }$ 
A) $\mathbf { r } ( u , v ) = u \mathbf { i } - v \mathbf { j } - ( 4 - u - v ) \mathbf { k }$
B) $\mathbf { r } ( u , v ) = u \mathbf { i } + v \mathbf { j } + ( 4 - u - v ) \mathbf { k }$
C) $\mathbf { r } ( u , v ) = u \mathbf { i } + v \mathbf { j } + ( 4 + u + v ) \mathbf { k }$
D) $\mathbf { r } ( u , v ) = u \mathbf { i } + v \mathbf { j } - ( 4 + u + v ) \mathbf { k }$
E) $\mathbf { r } ( u , v ) = - u \mathbf { i } - v \mathbf { j } + ( 4 - u - v ) \mathbf { k }$

Quizplus · Accepted Answer

Let's rewrite the equation of the plane as: $z=4-x-y$. Therefore, our vector-valued function should have the form: $\mathbf { r } ( u , v ) = a \mathbf { i }+ b \mathbf { j } + (4-a-b )\mathbf { k }$, where $a$ and $b$ can be variables in terms of $u$ and $v$ that will parameterize our function. 
Comparing this equation with choices A, B, C, D, and E, we can see that the correct answer is B because it has the form we are looking for with $a=u$, $b=v$, and $(4-a-b)=(4-u-v)$.

Find a Vector-Valued Function Whose Graph Is the Plane x+y+z=4. x + y + z = 4 \text {. }x+y+z=4.

Find a Vector-Valued Function Whose Graph Is the Plane $x + y + z = 4 \text {. }$