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Answer:
What is the magnitude and direction of the PQ−→− with tail and head points P(-6, 0) and Q(2, 4)? magnitude = √(64+16) = √80 = appr 8.9 , ruling out c.
hope this helps
The magnitude of PQ is 8.9 and PQ is making an angle of 26.57° so 8.9 units, 26.6° north of east will be the correct answer.
A vector is a quantity in which direction and magnitude both matters called a vector.
A vector joining by two points (x₁,y₁) and (x₂,y₂) will be (x₂-x₁)[tex]\hat{i}[/tex] + (y₂-y₁)[tex]\hat{j}[/tex]
The magnitude of this vector will be
[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2[/tex]
The direction of this vector will be
[tex]\\tan^{-1}(\frac{y_2-y_1}{x_2-x_1)}[/tex]
Given P(-6,0) and Q(2,4) so vector joining by
[tex]\vec{PQ}[/tex] = [2-(-6)][tex]\vec{i}[/tex] + [4-0][tex]\vec{j}[/tex]
[tex]\vec{PQ}[/tex] = 8[tex]\vec{i}[/tex] + 4[tex]\vec{j}[/tex]
now, the magnitude [tex]\vec{PQ}[/tex] will be
⇒ [tex]\sqrt{(8^2+4^2)}[/tex]
⇒ 8.944
The direction of the vector [tex]\vec{PQ}[/tex] will be
⇒ [tex]\tan^{-1}(\frac{4}{8})[/tex]
⇒26.57°
So, the magnitude of PQ is around 8.944 and the direction is 26.57° from the positive x axis so it's north of the east.
For more information about vector
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