# Thread: Finding a component of a coplanar vector of two other vectors

1. ## Finding a component of a coplanar vector of two other vectors

The question is determine the value of x for which the vector c = xi + j + k is coplanar with the vectors a ( i +j -k) and B (2i - j + 3k).
The anser is x = 4 and i think u need to use vector triple cross product to find it but i cant seem to manage it or find any examples of this style of question. Its seems to be a common style of question on most of the past papers of which my exam is tomoz.

2. Hello,
Originally Posted by mr_sputnik
The question is determine the value of x for which the vector c = xi + j + k is coplanar with the vectors a ( i +j -k) and B (2i - j + 3k).
The anser is x = 4 and i think u need to use vector triple cross product to find it but i cant seem to manage it or find any examples of this style of question.

Since vectors a and b are not colinear, they define a plane.

If you make the cross product between them two, you will get a vector d perpendicular to them two, that is, by extension, a vector orthogonal to the plane they form.

Thus, once you have the coordinates of d, find vector c such that its scalar product with d is 0, because any vector in this plane will be orthogonal to d.
Since you already have 2 coordinates of c, it will be easy

3. ur an ace chears, got correct anser

4. Hello, John!

Determine the value of $\displaystyle x$ for which the vector $\displaystyle \vec{C} \:= \:xi + j + k$
is coplanar with the vectors: $\displaystyle \vec{A} \:=\: i +j -k$ and $\displaystyle \vec{B} \:=\:2i - j + 3k$

The answer is $\displaystyle x = 4$
I think u need to use vector triple scalar product.

Recall that the triple scalar product, $\displaystyle \vec{C} \cdot (\vec{A} \times \vec{B})$, produces the volume
. . of the parallelepiped ('slanted box") determined by the three vectors.

If the three vectors are coplanar, the volume of the "box" is, of course, zero.

So we have: .$\displaystyle C\cdot(A \times B) \;=\;\left|\begin{array}{ccc}x & 1 & 1 \\ 1 & 1 & \text{-}1 \\ 2 & \text{-}1 & 3 \end{array}\right| \;=\;x(3-1)-(3+2) + (\text{-}1-2) \;=\;2x-8$

Therefore: .$\displaystyle 2x - 8 \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:4}$

5. if u read what i put more carefully rather than quoting me wroung you would see i put vector triple product ( a x (b x c) )with is differnt from the scalar triple product (a . (B x C) that is used to vol of parralapipehead.

6. Hello, mr_sputnik!

if u read what i put more carefully rather than quoting me wroung you would see i put vector triple product ( a x (b x c) )with is differnt from the scalar triple product (a . (B x C) that is used to vol of parralapipehead.

I did read what you wrote very carefully. . I did the problem correctly,
. . gave you the proper procedure and the correct answer.

So save your scolding for someone who did something stupid, okay?

. . I assumed that you meant triple scalar product

. . because (if you read my work) that is the proper formula for the problem.

Tale a look at your vector triple cross product . . .

$\displaystyle A \times (B \times C) \;=\;\langle x,1,1\rangle \times \left|\begin{array}{ccc} i & j & k \\ 1 & 1 & \text{-}1 \\ 2 & \text{-}1 & 3 \end{array}\right| \;=\;\langle x,1,1\rangle \times \langle 2,-5,3\rangle$

. . . . . . $\displaystyle = \;\left|\begin{array}{ccc}i & j & k \\ x & 1 & 1 \\ 2 & \text{-}5 & \text{-}3 \end{array}\right| \;=\;\langle 2,\:3x+2,\:-5x-2\rangle$

So you have a vector . . . now what?

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### value of k for which three vectors are coplanar

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