Pre-Calculus - Determine if Two Vectors Are Parallel or Perpendicular

Find the dot product of the two vectors

and

$\frac{d-b}{c-a}$

Explanation

To find the dot product of two vectors,

find the product of the x-components,

find the product of the y-components,

and then find the sum.

Which of the following pairs of vectors are parallel?

$r=\left<\frac{\sqrt3}{2}, \frac{1}{2}\right>, s=\left<\sqrt3, 1\right>$

$r=\left<\frac{\sqrt3}{2}, \frac{1}{2}\right>, s=\left<2\sqrt3, 1\right>$

$r=\left<\frac{\sqrt3}{4}, \frac{1}{2}\right>, s=\left<3\sqrt3, 3\right>$

$r=\left<-\frac{\sqrt3}{2}, -\frac{1}{2}\right>, s=\left<\sqrt3,\frac{1}{2}\right>$

Explanation

For two vectors, and to be parallel, , for some real number .

The correct answer here is $r=\left<\frac{\sqrt3}{2}, \frac{1}{2}\right>, s=\left<\sqrt3, 1\right>$ because

$r=\frac{1}{2}s$ .

Thus making the vector parallel.

$\frac{1}{2}\left<\sqrt3, 1\right>=\left<\frac{\sqrt3}{2}, \frac{1}{2}\right>$

Find the dot product of the two vectors

and

Explanation

To find the dot product of two vectors,

find the product of the x-components,

find the product of the y-components,

and then find the sum.

$1\cdot 0+0\cdot1=0$

$\vec{v}=-4i+5j$

$\vec{w}=5i+4j$

Which of the following best explains whether the two vectors above are perpendicular or parallel?

Perpendicular, because their dot product is zero.

Parallel, because their dot product is one.

Neither perpendicular nor parallel, because their dot product is neither zero nor one.

Perpendicular, because their dot product is one.

Parallel, because their dot product is zero.

Explanation

Two vectors are perpendicular if their dot product is zero, and parallel if their dot product is 1.

Take the dot product of our two vectors to find the answer:

$v_i\cdot w_i+v_j\cdot w_j=v \cdot w$

Using our given vectors:

$\vec{v}=-4i+5j$

$\vec{w}=5i+4j$

$v\cdot w=(-4\cdot 5)+(5\cdot 4)=-20+20=0$

Thus our two vectors are perpendicular.

The dot product may be used to determine the angle between two vectors.

Use the dot product to determine if the angle between the two vectors.

$\theta=6.667^o$

$\theta=81.90^o$

$\theta=69.54^o$

$\theta=21.76^o$

$\theta=86.30^o$

Explanation

First, we note that the dot product of two vectors is defined to be;

$a\cdot b=\left | a\right |\left | b\right |\cos\theta$ .

First, we find the left side of the dot product:

$a\cdot b= (5)(1)+(24)(3)=77$ .

Then we compute the lengths of the vectors:

$\begin{align*} \left | a\right |&= \sqrt{5^2+24^2}=\sqrt{601}\ \left | b\right |&= \sqrt{1^2+3^2}=\sqrt{10} \end{align*}$ .

We can then solve the dot product formula for theta to get:

$\cos^{-1}\frac{a \cdot b}{\left | a\right |\left | b\right |}=\theta$

Substituting the values for the dot product and the lengths will give the correct answer.

$\theta=6.667^o$

Find the angle between the following two vectors in 3D space.

$136.2^\circ$

$56.2^\circ$

$43.8^\circ$

$123.8^\circ$

$114.5^\circ$

Explanation

We can relate the dot product, length of two vectors, and angle between them by the following formula:

$a\cdot b=|a| \cdot |b| \cdot \cos \theta$

So the dot product of $a \cdot b$

and

is the addition of each product of components:

$2 \cdot -3 + 9 \cdot -4 + -3 \cdot 8 = -6 + (-36) + (-24) = -66$

now the length of the vectors of a and b can be found using the formula for vector magnitude:

$|a| = \sqrt{2^2+ 9^2 + (-3)^2} = \sqrt{94}$

$|b| = \sqrt{(-3)^2+ (-4)^2 + 8^2} = \sqrt{89}$

So:

$-66 = \sqrt94 \cdot \sqrt 89 \cdot \cos\theta$

$\cos \theta = -0.72158$ hence $\theta=136.2^\circ$

Which pair of vectors represents two parallel vectors?

$\left<6,3\right>, \left<2,1\right>$

$\left<1,3\right>,\left<2,4\right>$

$\left<6,3\right>,\left<4,8\right>$

$\left<1,1\right>,\left<2,1\right>$

$\left<11,3\right>,\left<33,3\right>$

Explanation

Two vectors are parallel if their cross product is . This is the same thing as saying that the matrix consisting of both vectors has determinant zero.

$\ \vector{v_1}&=\left<x_{1},y_{1}\right>\ \vector{v_2}&=\left<x_{2},y_{2}\right>\ M =\begin{pmatrix} x_{1}& y_{1}\ x_{2}&y_{2} \end{pmatrix}\det(M)=x_{1}y_{2}-x_{2}y_{1}=0\Rightarrow\vector{v_1}\parallel\vector{v_2}$