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GRE Subject Test: Math : GRE Subject Test: Math

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Example Questions

Example Question #91 : Vectors

Given points (0,8,-1) and (2,2,2) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 2,6,3\right \rangle$

$\left \langle 2,-6,-3\right \rangle$

$\left \langle -2,6,3\right \rangle$

$\left \langle 2,-6,3\right \rangle$

$\left \langle 2,6,-3\right \rangle$

Correct answer:

$\left \langle 2,-6,3\right \rangle$

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 (0,8,-1) and (2,2,2) , we get:

$v=\left \langle 2-0,2-8,2-(-1)\right \rangle$

$v=\left \langle 2,-6,3\right \rangle$

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Example Question #91 : Vector Form

What is the vector form of -6i+j+k ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 -6i+j+k , we can derive the vector form $\left \langle -6,1,1\right \rangle$ .

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Example Question #91 : Vector Form

What is the vector form of 3i-k ?

Possible Answers:

$\left \langle 3,0,-1\right \rangle$

$\left \langle 3,-1,0\right \rangle$

$\left \langle -3,0,-1\right \rangle$

$\left \langle 3,0,1\right \rangle$

$\left \langle -3,0,1\right \rangle$

Correct answer:

$\left \langle 3,0,-1\right \rangle$

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 3i-k , we can derive the vector form $\left \langle 3,0,-1\right \rangle$ .

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Example Question #113 : Linear Algebra

Given points (4,2,-2) and (7,1,0) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 3,1,2\right \rangle$

$\left \langle -3,1,-2\right \rangle$

$\left \langle 3,-1,2\right \rangle$

$\left \langle -3,1,2\right \rangle$

$\left \langle 3,1,-2\right \rangle$

Correct answer:

$\left \langle 3,-1,2\right \rangle$

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.

Subbing in our original points (4,2,-2) and (7,1,0) , we get:

$v=\left \langle 7-4,1-2,0-(-2)\right \rangle$

$v=\left \langle 3,-1,2\right \rangle$

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Example Question #91 : Vector Form

Given points (2,7,-6) and (1,2,3) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 1,-5,-9\right \rangle$

$\left \langle -1,5,-9\right \rangle$

$\left \langle -1,-5,-9\right \rangle$

$\left \langle -1,-5,9\right \rangle$

$\left \langle 1,5,9\right \rangle$

Correct answer:

$\left \langle -1,-5,9\right \rangle$

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.

Subbing in our original points (2,7,-6) and (1,2,3) , we get:

$v=\left \langle 1-2,2-7,3-(-6)\right \rangle$

$v=\left \langle -1,-5,9\right \rangle$

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Example Question #91 : Vectors

What is the vector form of -i+2j+10k ?

Possible Answers:

$\left \langle 1,2,-10\right \rangle$

$\left \langle 1,-2,10\right \rangle$

$\left \langle 1,2,10\right \rangle$

$\left \langle -1,2,10\right \rangle$

$\left \langle 1,-2,-10\right \rangle$

Correct answer:

$\left \langle -1,2,10\right \rangle$

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 -i+2j+10k , we can derive the vector form $\left \langle -1,2,10\right \rangle$ .

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Example Question #92 : Vector Form

What is the vector form of i+k ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 i+k , we can derive the vector form $\left \langle 1,0,1\right \rangle$ .

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Example Question #91 : Vectors

Calculate the dot product of the following vectors:

$a=\left \langle c,6, s\right \rangle$

$b=\left \langle c,10,s\right \rangle$

Possible Answers:

c^2+60+s^2

c^2+16+s^2

2c+16+2s

2cs+60

60c^2s^2

Correct answer:

c^2+60+s^2

Explanation:

Write the formula for dot product given $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_{1},b_{2},b_{3}\right \rangle$ .

$a\cdot b= a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}$

Substitute the values of the vectors to determine the dot product.

$a\cdot b= c(c)+(6)(10)+s(s) = c^2+60+s^2$

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Example Question #1 : Solving Systems Of Equations

2y=6x+4

y=7x-2

Solve the system of equations.

Possible Answers:

x=4, y=8

$x=-8, y=\frac{1}{2}$

x=1, y=5

x=-1, y=3

x=-2, y=-3

Correct answer:

x=1, y=5

Explanation:

The easiest way to solve this question is to use substitution. Since y=7x-2 you can replace y for 7x-2 in the other equation.

You should have

2(7x-2)=6x+4 .

Distribute the 2 to the parentheses.

14x-4=6x+4

Add 4 to both sides of the equation.

14x=6x+8

Subtract 6x from both sides.

8x=8

Divide by 8 to get x.

x=1

Put 1 back in to either equation for x to solve for y.

y=3(1)+2

y=5

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Example Question #2 : Solving Systems Of Equations

-3x+2y=19

-3x-19=2y

Solve the system of equations.

Possible Answers:

$x=-6\frac{1}{3}, y=0$

$x=\frac{1}{4}, y=-2\frac{1}{5}$

$x=3, y=-4\frac{1}{5}$

$x=-1, y=-4\frac{1}{3}$

$x=-\frac{1}{5}, y=2\frac{1}{4}$

Correct answer:

$x=-6\frac{1}{3}, y=0$

Explanation:

First task is to solve at least one of the equations for y.

-3x+2y=19

Move -3x to the other side by adding 3x to both sides.

2y=3x+19

Divide by 2 to all the terms in the equation.

$y=\frac{3}{2}x+\frac{19}{2}$

Plug this value for y into the other equation.

$-3x-19=2(\frac{3}{2}x+\frac{19}{2})$

Distribute the 2.

-3x-19=3x+19

Add 3x to both sides.

-19=6x+19

Subtract 19 from both sides of the equation.

-38=6x

Divide by 6.

$x=-6.33 or -6\frac{1}{3}$

Plug this back in for x in either equation.

-3(-6.33)+2y=19

19+2y=19

2y=0

y=0

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GRE Subject Test: Math : GRE Subject Test: Math

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #91 : Vectors

Example Question #91 : Vector Form

Example Question #91 : Vector Form

Example Question #113 : Linear Algebra

Example Question #91 : Vector Form

Example Question #91 : Vectors

Example Question #92 : Vector Form

Example Question #91 : Vectors

Example Question #1 : Solving Systems Of Equations

Example Question #2 : Solving Systems Of Equations

All GRE Subject Test: Math Resources

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