Question

If $u,v,w$ are non - coplanar vectors and p, q are real numbers, then the equality $\left[3upvpw\right]–\left[pvwqu\right]–\left[2wqvqu\right]=0$ holds for?

• A. exactly two values of $\left(p,q\right)$
• B. more than two but not all values of $(p,q)$
• C. all values of$(p,q)$
• D. exactly one value of $(p,q)$

Answer 1

The correct option is D

exactly one value of $(p,q)$

Explanation for correct option:

Finding the relation:

Given that,

$u,v,w$are non–coplanar vectors and $p,q$ are real numbers.

Simplifying the above equation, we get

$$\begin{gathered} \left[3upvpw\right]–\left[pvwqu\right]–\left[2wqvqu\right]=0 \\ \Rightarrow 3p^2\left[u.(vxw)\right]–pq\left[v.(wxu)\right]–2q^2\left[w.(vxw)\right]=0 \\ \Rightarrow \left(3p^2-pq+2q^2\right)\left[u.(vxw)\right]=0 \end{gathered}$$

Since$u,v,w$ are non - coplanar vectors

$\left[uvw\right]\ne 0$

so, $(3p^2-pq+2q^2)=0$

or $p=q=0$

Therefore, there is only one value of $p,q$.

Hence, the correct option is D.