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

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

Grade:11

5 Answers

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

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