Question

Find the vector having magnitude √6 units and perpendicular to both 2i - k and 3j - i - k.

Answer 1

Answer:

w = - i - j -2 k     is the Required Vector.

Step-by-step explanation:

The given vectors are

u = 2i + 0j - k

v = 3j - i - k

Now

u × v = $\left[\begin{array}{ccc}i&j&k\\2&0&-1\\3&-1&-1\end{array}\right]$

u × v = i (0 - 1 ) - j ( - 2 + 3 ) + k ( - 2 - 0 )

u × v = - i - j -2 k

|u × v| = $\sqrt{(-1)^{2} +(-1)^{2}+(-2)^{2}}=\sqrt{6}$

Now

Let n be the required unit vector Perpendicular to this Plane, then

n = [u × v] / |u × v|

n = $\frac{1}{\sqrt{6} } [- i - j -2 k]$

Now let w be the required vector, then

w  = √6  n

$W=\sqrt{6}\frac{1}{\sqrt{6} } [- i - j -2 k]$

w = - i - j -2 k       is the Required Vector.

Answer 2

Hence, this is the required result.

i hope my answer is correct