AP Calculus BC - Vector Form

What is the vector form of ?

$\left \langle 10,-1,10\right \rangle$

$\left \langle 10,10,-1\right \rangle$

$\left \langle -1,10,10\right \rangle$

$\left \langle 10,1,10\right \rangle$

$\left \langle 10,10,1\right \rangle$

Explanation

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for , we can derive the vector form $\left \langle 10,-1,10\right \rangle$ .

What is the vector form of ?

$\left \langle -6,1,1\right \rangle$

$\left \langle 6,-1,1\right \rangle$

$\left \langle 6,1,-1\right \rangle$

$\left \langle -6,1,-1\right \rangle$

$\left \langle 6,1,1\right \rangle$

Explanation

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for , we can derive the vector form $\left \langle -6,1,1\right \rangle$ .

What is the vector form of ?

$\left \langle 7,-4,1\right \rangle$

$\left \langle 1,7,-4\right \rangle$

$\left \langle7,-4,1\right \rangle$

$\left \langle-4,1,7\right \rangle$

$\left \langle1,-4,7\right \rangle$

Explanation

Given , we need to map the , , and coefficients back to their corresponding , , and -coordinates.

Thus the vector form of is

$\left \langle 7,-4,1\right \rangle$ .

Given points and , what is the vector form of the distance between the points?

$\left \langle 3,2,-8\right \rangle$

$\left \langle -3,2,-8\right \rangle$

$\left \langle -3,-2,-8\right \rangle$

$\left \langle -3,2,8\right \rangle$

$\left \langle 3,2,8\right \rangle$

Explanation

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points. That is, for any point $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , the distance is the vector $v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points and , we get:

$v=\left \langle 5-2,1-(-1),-1-7\right \rangle$

$v=\left \langle 3,2,-8\right \rangle$

What is the vector form of ?

$\left \langle 4,-7,2\right \rangle$

$\left \langle 4,-7,-2\right \rangle$

$\left \langle -4,-7,-2\right \rangle$

$\left \langle 4,7,2\right \rangle$

$\left \langle -4,7,-2\right \rangle$

Explanation

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for , we can derive the vector form $\left \langle 4,-7,2\right \rangle$ .

What is the vector form of ?

$\left \langle 0,1,0\right \rangle$

$\left \langle 0,-1,0\right \rangle$

$\left \langle 0,0,1\right \rangle$

$\left \langle 1,0,0\right \rangle$

$\left \langle -1,0,0\right \rangle$

Explanation

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for , we can derive the vector form $\left \langle 0,1,0\right \rangle$ .

What is the vector form of ?

$\left \langle -1,1,1\right \rangle$

$\left \langle 1,-1,1\right \rangle$

$\left \langle 1,1,-1\right \rangle$

$\left \langle -1,-1,1\right \rangle$

$\left \langle 1,-1,1\right \rangle$

Explanation

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients. That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ . So for , we can derive the vector form $\left \langle -1,1,1\right \rangle$ .

What is the vector form of ?

$\left \langle -2,9,0\right \rangle$

$\left \langle 0,-2,9\right \rangle$

$\left \langle9, 0,-2\right \rangle$

$\left \langle-2, 0, 9\right \rangle$

None of the above

Explanation

In order to derive the vector form, we must map the vector elements to their corresponding , , and coefficients. That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ . So for , we can derive the vector form $\left \langle -2,9,0\right \rangle$ .