Vectors question/confusion

**IanCarney** · May 6th 2011, 01:29 PM

Determine the value for $x$ such that the point A(-1,3,4), B(-2,3,-1), and C(-5,6, $x$ ) 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.

**VincentP** · May 6th 2011, 01:37 PM

Try using this:
If 3 vectors lie on the same plane their Triple product is 0.

**mtpastille** · May 6th 2011, 01:42 PM

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 $\overrightarrow{OA}$ and $\overrightarrow{OB}$ . Then find a Cartesian equation of this plane (i.e., in the form $Ax+By+Cz=D$ ). 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

**Plato** · May 6th 2011, 01:55 PM

Originally Posted by IanCarney

Determine the value for $x$ such that the point A(-1,3,4), B(-2,3,-1), and C(-5,6, $x$ ) 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.

Here is another way. Think of $A,~B,~\&~C$ as vectors.
The solve $C\cdot(A\times B)=0$ .

**Soroban** · May 6th 2011, 06:41 PM

Hello, IanCarney!

Here is an elementary solution . . .

Determine the value for $\,a$ such that the points $A(\text{-}1,3,4),\;B(\text{-}2,3,\text{-}1),\;C(\text{-}5,6,a)$
all lie on a plane that contains the origin. . (Answer: -7)

A plane has the general equation: . $Ax + By + Cz + D \:=\:0$
. . If the plane contains the Origin, then $D \,=\,0.$

$\text{Let }\vec u \:=\:\overrightarrow{BA} \:=\:\langle 1,0,5\rangle$
$\text{Let }\vec v \:=\:\overrightarrow{BC} \:=\:\langle \text{-}3,3,a+1\rangle$

$\text{The normal to the plane is: }\:\vec n \;=\;\vec u \times \vec v \;=\;\begin{vmatrix}i & j & k \\ 1 & 0 & 5 \\ \text{-}3 & 3 & a\!+\!1 \end{vmatrix}$

. . $\vec n \;=\;i(0\!-\!15) - j(a\!+\!1\!+\!15) + k(3\!-\!0) \;=\;\text{-}15i - (a\!+\!16)j + 3k$

. . . . $=\;\langle \text{-}15,\,\text{-}(a\!+\!16),\:3\rangle \;=\;\langle 15,\:a\!+\!16,\:\text{-}3\rangle$

The equation of the plane through $A(\text{-}1,3,4)$ with $\vec n \,=\,\langle 15, a\,+\,\!1,\text{-}3\rangle$ is:

. . . . . . $15(x+1) + (a+16)(y-3) - 3(z - 4) \;=\;0$

. . . $15x + 15 + (a+16)y - 3(a+16) - 3z + 12 \;=\;0$

. . . . . . . . . . . $15x + (a+16)y - 3z \underbrace{-\:3a - 21}_{\text{This is }D} \;=\;0$

Therefore: . $-3a - 21 \:=\:0 \quad\Rightarrow\quad \boxed{a \:=\:-7}$

Thread: Vectors question/confusion

LinkBack

Thread Tools

Display

Vectors question/confusion

Similar Math Help Forum Discussions

[SOLVED] Inequalities and Surds - dying of confusion over this question.

Sum question,some confusion

Taylor series question confusion

Utter confusion over a simple question

find angle between vectors .. confusion.

Search Tags