Question

Given that P = (-5, 11) and Q = (-6, 4), find the component form and magnitude of vector QP.

<-11, 7>, square root of 38
<-1, -7>, square root of 50
<1, 7>, 50
<1, 7>, square root of 50

Answer 1

We'll use the Pythagorean Theorem (otherwise known as the Distance Formula):

Going from Q to P, x increases by 1 and y increases by 7. Thus, the component form of QP is <1,7>, and the magnitude of this vector is √(1^2+7^2) = 5√2.

Note: this is equivalent to <1, 7>, square root of 50.

Answer 2

Answer with explanation:

P=(-5,11)=-5 i +11 j

Q=(-6,4)= -6 i+ 4 j

⇒Component of Vector QP

=Position vector of P -Position vector of Q

= -5 i + 11 j - (-6 i+4 j)

= -5 i +11 j +6 i -4 j

= i +7 j------------Adding like terms

⇒Magnitude of vector QP

`=√((1)^2+(7)^2)\\\\=√(50)`

Option D