Question

_ _ _ _ _ _ If x=(3,1,0),y = (2,2,3), z= (-1,2,1) and x ⊥(y + kz), then K=.......,Select correct option from the given options. (a) 8 (b) 4 (c) 1/8 (d) 1/4

Answer 1

Answer:

K= 8

Step-by-step explanation:

In this question,

we have been given that

if x=(3,1,0)

  y=(2,2,3)

  z=(-1,2,1) and x is perpendicular to (y+kz)

Then we need to find the value of k

$x\ =\ 3\hat{i}+\hat{j}$

$y\ =\ 2\hat{i}+\hat{k}+3\hat{j}$

$z\ =\ -\hat{i}+2\hat{j}+\hat{k}$

Since, x is perpendicular to (y+kz)

Therefore y+kz = $(2-k)\hat{i}+(2+2k)\hat{j}+(3+k)\hat{k}$

Therefore, using dot product we get,

(x).(y+kz) = (x).(y+kz) cosθ

Here θ = 90

Therefore (x).(y+kz) = (x).(y+kz) cos(90)      

$(3\hat{i}+\hat{j}).(2-k)\hat{i}+(2+2k)\hat{j}+(3+k)\hat{k}$ = 0     Since {cos90 = 0}

(2-k)3+(2+2k)1 = 0

6 - 3k + 2 + 2k = 0

k = 8

Hence value of K is 8