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Solution for the given ...

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Asked by joshuasimon2252003 1st June 2020, 10:42 PM
Answered by Expert
Answer:
Given: vector a = i + 2j - 3k  and  vector b = 3i - j +2k
To show that (vector a + vector b) is perpendicular to (vector a - vector b)
Two vectors are perpendicular to each other if and only if their dot product is zero.
Consider space the space dot space product space of space open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses space and space open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses
open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses times open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses equals straight a with rightwards arrow on top times straight a with rightwards arrow on top minus straight a with rightwards arrow on top times straight b with rightwards arrow on top plus straight b with rightwards arrow on top times straight a with rightwards arrow on top minus straight b with rightwards arrow on top times straight b with rightwards arrow on top
equals straight a with rightwards arrow on top times straight a with rightwards arrow on top minus straight a with rightwards arrow on top times straight b with rightwards arrow on top plus straight a with rightwards arrow on top times straight b with rightwards arrow on top minus straight b with rightwards arrow on top times straight b with rightwards arrow on top space space... space open parentheses Since space straight a with rightwards arrow on top times straight b with rightwards arrow on top equals straight b with rightwards arrow on top times straight a with rightwards arrow on top close parentheses
equals straight a with rightwards arrow on top times straight a with rightwards arrow on top minus straight b with rightwards arrow on top times straight b with rightwards arrow on top
equals open vertical bar straight a with rightwards arrow on top close vertical bar squared minus open vertical bar straight b with rightwards arrow on top close vertical bar squared
equals square root of 1 squared plus 2 squared plus open parentheses negative 3 close parentheses squared end root minus square root of 3 squared plus open parentheses negative 1 close parentheses squared plus 2 squared end root
equals square root of 14 minus square root of 14
equals 0
Therefore comma space open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses times open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses equals 0
Hence comma space straight a with rightwards arrow on top plus straight b with rightwards arrow on top space is space perpendicular space to space straight a with rightwards arrow on top minus straight b with rightwards arrow on top.
Answered by Expert 2nd June 2020, 10:49 AM
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