Determine the value for $\displaystyle x$ such that the point A(-1,3,4), B(-2,3,-1), and C(-5,6,$\displaystyle x$) all lie on a plane that contains the origin.
I'm not sure what they're asking for. The answer in the book is -7.
Determine the value for $\displaystyle x$ such that the point A(-1,3,4), B(-2,3,-1), and C(-5,6,$\displaystyle x$) all lie on a plane that contains the origin.
I'm not sure what they're asking for. The answer in the book is -7.
You have four points: A, B, C and O (origin). The question is asking for a plane, so try to find an equation of a plane that involves all of these four points. Start by constructing two vectors from these points, say $\displaystyle \overrightarrow{OA}$ and $\displaystyle \overrightarrow{OB}$. Then find a Cartesian equation of this plane (i.e., in the form $\displaystyle Ax+By+Cz=D$). Since the plane intersects the origin, D=0. See if you can then solve the equation for x.
Edit: I suppose VincentP's method is more sophisticated, but both work
Hello, IanCarney!
Here is an elementary solution . . .
Determine the value for $\displaystyle \,a$ such that the points $\displaystyle A(\text{-}1,3,4),\;B(\text{-}2,3,\text{-}1),\;C(\text{-}5,6,a)$
all lie on a plane that contains the origin. . (Answer: -7)
A plane has the general equation: .$\displaystyle Ax + By + Cz + D \:=\:0$
. . If the plane contains the Origin, then $\displaystyle D \,=\,0.$
$\displaystyle \text{Let }\vec u \:=\:\overrightarrow{BA} \:=\:\langle 1,0,5\rangle$
$\displaystyle \text{Let }\vec v \:=\:\overrightarrow{BC} \:=\:\langle \text{-}3,3,a+1\rangle$
$\displaystyle \text{The normal to the plane is: }\:\vec n \;=\;\vec u \times \vec v \;=\;\begin{vmatrix}i & j & k \\ 1 & 0 & 5 \\ \text{-}3 & 3 & a\!+\!1 \end{vmatrix}$
. . $\displaystyle \vec n \;=\;i(0\!-\!15) - j(a\!+\!1\!+\!15) + k(3\!-\!0) \;=\;\text{-}15i - (a\!+\!16)j + 3k $
. . . . $\displaystyle =\;\langle \text{-}15,\,\text{-}(a\!+\!16),\:3\rangle \;=\;\langle 15,\:a\!+\!16,\:\text{-}3\rangle $
The equation of the plane through $\displaystyle A(\text{-}1,3,4)$ with $\displaystyle \vec n \,=\,\langle 15, a\,+\,\!1,\text{-}3\rangle$ is:
. . . . . . $\displaystyle 15(x+1) + (a+16)(y-3) - 3(z - 4) \;=\;0$
. . .$\displaystyle 15x + 15 + (a+16)y - 3(a+16) - 3z + 12 \;=\;0$
. . . . . . . . . . . $\displaystyle 15x + (a+16)y - 3z \underbrace{-\:3a - 21}_{\text{This is }D} \;=\;0$
Therefore: .$\displaystyle -3a - 21 \:=\:0 \quad\Rightarrow\quad \boxed{a \:=\:-7}$