If#" "veca=3hati+4hatj+5hatk and vec b= 2hati+hatj-4hatk# ;How will you find out the component of #" "veca " ""perpendicular to" " " vecb#?

2 Answers
Mar 6, 2017

Answer:

#(83/21, 94/21, 65/21)#

Explanation:

Let #vec a = vec a_1 + vec a_2# with #vec a_1 ne vec 0, vec a_2 ne vec 0#such that #<< vec a_1, vec a_2 >> = 0# and # vec a_2 = lambda vec b# with #lambda in RR#

Making now #<< vec a, vec b >> = << vec a_1+vec a_2 , vec b>> = << vec a_1, vec b >> + << vec a_2, vec b >># but by hypothesis # vec a_2 = lambda vec b# so

#<< vec a, vec b >> = lambda << vec b, vec b >># so

#lambda = (<< vec a, vec b >>)/(<< vec b, vec b >>)# so

#vec a_2 = (<< vec a, vec b >>)/(<< vec b, vec b >>) vec b# and

#vec a_1 = vec a - (<< vec a, vec b >>)/(<< vec b, vec b >>) vec b#

or

#vec a_1 = (3,4,5)-(3 xx 2+4 xx 1-5 xx 4)/(2^2+1^2+4^2)(2,1,-4) = (83/21, 94/21, 65/21)#

Mar 6, 2017

The component of #veca# perpendicular to #vecb#

#=veca-"projection of "veca" " on " " vecb#

#=veca-(veca*vecb)/abs(vecb)xxvecb/abs(vecb)#

#=veca-(veca*vecb)/abs(vecb)^2xxvecb#

#=(3hati+4hatj+5hatk)-((3hati+4hatj+5hatk).(2hati+hatj-4hatk))/abs(2hati+hatj-4hatk)^2xx(2hati+hatj-4hatk)#

#=(3hati+4hatj+5hatk)-(2*3+4*1+5*-4)/(2^2+1^2+4^2)xx(2hati+hatj-4hatk)#

#=(3hati+4hatj+5hatk)+10/21xx(2hati+hatj-4hatk)#

#=1/21[21(3hati+4hatj+5hatk)+10(2hati+hatj-4hatk)]#

#=1/21[83hati+94hatj+65hatk]#