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# If 3i^+2j^+8k^and 2i^+xj^+k^ are at right angle then x

Arun
25763 Points
3 years ago
Dear student

if these are at right angles then their dot product should be zero.

Hence

(3i + 2j + 8k) . (2i + xj + k) = 0

6 + 2x + 8 = 0

2x = – 14

x = – 7
Khimraj
3007 Points
3 years ago
for two vector at right angle their dot product will be zero as cos90 is zero
So
(3$\hat{i}$ +2$\hat{j}$ + 8$\hat{k}$).(2$\hat{i}$ + x$\hat{j}$ + $\hat{k}$) = 0
then
6 + 2x + 8 = 0
x = -7
Hope it clears.
Tony
108 Points
2 years ago
Whenever two vectors are perpendicular then take their dot product as zero.  Hope it helps. X= - 7 by dot product!
Naveen sharma
16 Points
2 years ago

(3$\hat{i}$ +2$\hat{j}$ + 8$\hat{k}$).(2$\hat{i}$ + x$\hat{j}$ + $\hat{k}$) = 0
then
6 + 2x + 8 = 0
x = -7 There is right angle triangle so angel of 90 is made so cos90=0
Kaviya
17 Points
2 years ago
If the two vectors are at 90° then their dot product will be equal to zero.
A=3i+2j+8k
B=2i+xj+k

So, A.B=0
(3i+2j+8k).(2i+xj+k)=0
6+2x+8=0
Therefore X=-7