If a,b,c are 3 mutually perpendicular vectors of the same magnitude, prove that a+b+c is equally inclined with vectors a,b and c.

Asked by sunil2791 | 21st Nov, 2017, 06:00: PM

Expert Answer:

If three vectors are mutually perpendicular,
begin mathsize 16px style straight a with rightwards harpoon with barb upwards on top times straight b with rightwards harpoon with barb upwards on top equals straight b with rightwards harpoon with barb upwards on top times straight c with rightwards harpoon with barb upwards on top equals straight c with rightwards harpoon with barb upwards on top times straight a with rightwards harpoon with barb upwards on top equals 0
Also space according space to space the space question space
open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar equals open vertical bar straight b with rightwards harpoon with barb upwards on top close vertical bar equals open vertical bar straight c with rightwards harpoon with barb upwards on top close vertical bar equals straight lambda
open vertical bar straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top close vertical bar squared equals open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared plus open vertical bar straight b with rightwards harpoon with barb upwards on top close vertical bar squared plus open vertical bar straight c with rightwards harpoon with barb upwards on top close vertical bar squared plus 2 open parentheses straight a with rightwards harpoon with barb upwards on top times straight b with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top times straight c with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top times straight a with rightwards harpoon with barb upwards on top close parentheses
open vertical bar straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top close vertical bar squared equals open vertical bar straight a with rightwards harpoon with barb upwards on top close vertical bar squared plus open vertical bar straight b with rightwards harpoon with barb upwards on top close vertical bar squared plus open vertical bar straight c with rightwards harpoon with barb upwards on top close vertical bar squared
open vertical bar straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top close vertical bar squared equals 3 straight lambda squared
open vertical bar straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top close vertical bar equals square root of 3 straight lambda
Now space find space the space angle space between space straight a with rightwards harpoon with barb upwards on top and space straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top space which space gives space you
cos to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses space.
Similarly space you space can space show space that space straight b with rightwards harpoon with barb upwards on top and space straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top space is space cos to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses space.
Also space you space can space show space that space straight c with rightwards harpoon with barb upwards on top and space straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top space is space cos to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses space.
Hence comma space straight a with rightwards harpoon with barb upwards on top plus straight b with rightwards harpoon with barb upwards on top plus straight c with rightwards harpoon with barb upwards on top space is space equally space inclined space with space straight a with rightwards harpoon with barb upwards on top comma straight b with rightwards harpoon with barb upwards on top comma straight c with rightwards harpoon with barb upwards on top.
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Answered by Sneha shidid | 29th Nov, 2017, 11:40: AM