Find the x that makes <2,x> and <4,-6> parallel?

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Answer 1 · 2018-08-09T08:42:29

#x = -8#

Recall..

For #x# to be parallel it means the gradients must be equal to #1#

#m = 1#

#m_1 = (y_2 - y_1)/(x_2 - x_1)#

#y_2 = -6#

#y_1 = x#

#x_2 = 4#

#x_1 = 2#

Plugging in the values..

#m_1 = ((-6) - x)/(4 - 2) = (-6 - x)/2#

Therefore;

#(-6 - x)/2 = 1#

#(-6 - x)/2 = 1/1#

Cross multiply;

#(-6 - x) xx 1 = 1 xx 2#

#-6 - x = 2#

#-x = 2 + 6#

#-x = 8#

#x = -8#

Answer 2 · 2018-08-09T09:25:42

#x=-3#.

By definition, two non-null vectors

#vecu, &, vecv" are parallel "iff vecu=kvecv" for some "k in RR-{0}#.

#:. (2,x)=k(4,-6) rArr 4k=2, -6k=x#.

#:. k=2/4=1/2#.

#:. x=-6k=-6*1/2=-3#.