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


2 years ago

Arun
24742 Points
							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

2 years ago
Khimraj
3007 Points
							for two vector at right angle their dot product will be zero as cos90 is zeroSo(3$\hat{i}$ +2$\hat{j}$ + 8$\hat{k}$).(2$\hat{i}$ + x$\hat{j}$ + $\hat{k}$) = 0then6 + 2x + 8 = 0x = -7Hope it clears.

2 years ago
Tony
108 Points
							Whenever two vectors are perpendicular then take their dot product as zero.  Hope it helps. X= - 7 by dot product!

2 years ago
Naveen sharma
16 Points
							 (3$\hat{i}$ +2$\hat{j}$ + 8$\hat{k}$).(2$\hat{i}$ + x$\hat{j}$ + $\hat{k}$) = 0then6 + 2x + 8 = 0x = -7 There is right angle triangle so angel of 90 is made so cos90=0

2 years ago
Kaviya
17 Points
							If the two vectors are at 90° then their dot product will be equal to zero.A=3i+2j+8kB=2i+xj+k So, A.B=0(3i+2j+8k).(2i+xj+k)=06+2x+8=0Therefore X=-7

2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions