Question

Let a ,b ,c be non zero non coplanar vectors . Prove that a-2b+3c , -2a+3b-4c and a-3b+5c are coplanar vectors .

Answer 1

hey mate
here's the solution

Answer 2

Answer:

To prove co-planar we need to prove  x · [y × z] = 0

Let x be a - 2b + 3c, y be -2a + 3b - 4c and z be a - 3b + 5c

$\begin{pmatrix}i&j&k\\ -2&3&-4\\ 1&-3&5\end{pmatrix}$

$i\det \begin{pmatrix}3&-4\\ -3&5\end{pmatrix}-j\det \begin{pmatrix}-2&-4\\ 1&5\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ 1&-3\end{pmatrix}$

$=1\cdot \left(-3\right)-2\left(-6\right)+3\left(-3\right)$

$=\left-3i\right+6j\left\right-9k\left$

Now, take dot product of a - 2b + 3c with $=\left-3i\right+6j\left\right-9k\left$

$\mathrm{Computing\:dot\:product\:of\:two\:vectors}:\quad \left(x_1,\:\:\ldots ,\:\:x_n\right)\cdot \left(y,\:\:\ldots ,\:\:y_n\right)=\sum _{i=1}^nx_iy_i$

$\begin{pmatrix}1&-2&3\end{pmatrix}\begin{pmatrix}-3&6&-3\end{pmatrix}$

$=1\cdot \left(-3\right)+\left(-2\right)\cdot \:6+3\left(-3\right)$

$=0$

Hence proved , a-2b+3c , -2a+3b-4c and a-3b+5c are coplanar vectors .