How do you find the exact value(s) of K?

Find the exact value(s) of K, if any, so that #k##vec u#+#3##vec v# is orthogonal to #vec z#=#<1,1,1>#

1 Answer
George C.
Aug 4, 2018

According to whether #vec(u)# and/or #vec(v)# are orthogonal to #vec(z)# there may be no, one or infinitely many possible values of #k#...

Explanation:

Two non-zero vectors are orthogonal if and only if their dot product is zero.

Given:

#vec(u) = < u_1, u_2, u_3 >#

#vec(v) = < v_1, v_2, v_3 >#

Then:

#(kvec(u)+3vec(v)) * vec(z)#

#= (k < u_1, u_2, u_3 > + 3 < v_1, v_2, v_3 > ) * < 1, 1, 1>#

#= < k u_1 + 3 v_1, k u_2 + 3 v_2, k u_3 + 3 v_3 > * < 1, 1, 1 >#

#= (k u_1 + 3 v_1) + (k u_2 + 3 v_2) + (k u_3 + 3 v_3)#

#= k(u_1+u_2+u_3)+3(v_1+v_2+v_3)#

If neither #vec(u)# nor #vec(v)# are orthogonal to #vec(z)# then both #u_1+u_2+u_3 != 0# and #v_1+v_2+v_3 != 0#, yielding one possible value for #k#, namely:

#k = -(3(v_1+v_2+v_3))/(u_1+u_2+u_3)#

If #vec(u)# is orthogonal to #vec(z)# then #u_1+u_2+u_3 = 0# and the value of #k# has no effect: Either #vec(v)# is orthogonal to #vec(z)# too and any value of #k# will work or #vec(v)# is not orthogonal to #vec(z)# and no value of #k# will work.

If #vec(u)# is not orthogonal to #vec(z)# but #vec(v)# is, then the only value of #k# which works is #k=0#.

Related questions
Impact of this question
1809 views around the world
You can reuse this answer
Creative Commons License