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GRE Subject Test: Math : Linear Algebra

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Example Questions

Example Question #434 : Parametric, Polar, And Vector

What is the vector form of $\displaystyle 10i-j+10k$?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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Example Question #84 : Vector

What is the vector form of i+j $\displaystyle i+j$?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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Example Question #91 : Vectors

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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

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

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Example Question #91 : Vector Form

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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Example Question #91 : Vector Form

What is the vector form of 3i-k $\displaystyle 3i-k$?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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Example Question #1 : Vector Form

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\displaystyle (4,2,-2)$ and (7,1,0) $\displaystyle (7,1,0)$, we get:

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

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

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Example Question #91 : Vector Form

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\displaystyle (2,7,-6)$ and (1,2,3) $\displaystyle (1,2,3)$, we get:

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

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

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Example Question #441 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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Example Question #92 : Vector Form

What is the vector form of i+k $\displaystyle i+k$?

Possible Answers:

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

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

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

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

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

Correct answer:

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

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

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Example Question #91 : Vectors & Spaces

Calculate the dot product of the following vectors:

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

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

Possible Answers:

$\displaystyle 60c^2s^2$

$\displaystyle c^2+60+s^2$

2cs+60 $\displaystyle 2cs+60$

$\displaystyle 2c+16+2s$

$\displaystyle c^2+16+s^2$

Correct answer:

$\displaystyle c^2+60+s^2$

Explanation:

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

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

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

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

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GRE Subject Test: Math : Linear Algebra

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

All GRE Subject Test: Math Resources

Example Questions

Example Question #434 : Parametric, Polar, And Vector

Example Question #84 : Vector

Example Question #91 : Vectors

Example Question #91 : Vector Form

Example Question #91 : Vector Form

Example Question #1 : Vector Form

Example Question #91 : Vector Form

Example Question #441 : Parametric, Polar, And Vector

Example Question #92 : Vector Form

Example Question #91 : Vectors & Spaces

All GRE Subject Test: Math Resources

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