Determine the value(s) of k such that (2k, 3, k+1) x (k-1, k, 1) = (1, -2, 2)?

Apr 28, 2017

$k = 1$

Explanation:

Cross product of $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ is given by

$\left({y}_{1} {z}_{2} - {y}_{2} {z}_{1} , {z}_{1} {x}_{2} - {z}_{2} {x}_{1} , {x}_{1} {y}_{2} - {y}_{1} {x}_{2}\right)$

Hence cross product of $\left(2 k , 3 , k + 1\right)$ and $\left(k - 1 , k , 1\right)$ is

((3xx1-k(k+1)),((k+1)(k-1)-1xx2k),(2kxxk-3(k-1))

or $\left(3 - {k}^{2} - k , {k}^{2} - 1 - 2 k , 2 {k}^{2} - 3 k + 3\right)$

As it is $\left(1. - 2.2\right)$, we have

$3 - {k}^{2} - k = 1$ i.e. ${k}^{2} + k - 2 = 0$ i.e. $k = - 2$ or $k = 1$ and

${k}^{2} - 2 k - 1 = - 2$ i.e. ${k}^{2} - 2 k + 1 = 0$ i.e. $k = 1$ and

$2 {k}^{2} - 3 k + 3 = 2$ i.e. $2 {k}^{2} - 3 k + 1 = 0$ i.e. $k = \frac{1}{2}$ or $k = 1$

As value of $k$ should hold for all components of vector.

we have $k = 1$