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Vectors Progress Test

Authored by Peter Lego

Mathematics

12th Grade

Used 3+ times

Vectors Progress Test
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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Let u = i - 2j - 3k and v = 2i + j - k. The vector resolute of u perpendicular to v is:

 −52-\frac{5}{2} (j + k)

-5(i + j)

-2i  -\frac{5}{2} j  -\frac{5}{2} k

i +  −12-\frac{1}{2} j  −12-\frac{1}{2} k 

2i +  \frac{1}{2} j  -\frac{1}{2} k

2.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

A unit vector in the opposite direction to 2i - 2j + k is:

-2i + 2j - k

 12\frac{1}{2}  (-2i + 2j - k)

 13\frac{1}{3}  (-2i + 2j - k)

 14\frac{1}{4}  (-2i + 2j - k)

 15\frac{1}{5}  (-2i + 2j - k)

3.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

The vectors a = i + j + 2k, b = -i + 2j - 2k, c = i + mk are linearly dependent, where m is a real constant. The value of m is:


−13-\frac{1}{3}

0

23\frac{2}{3}

1

2

4.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

The vectors a = 2i + j - k, b = mi + nj, and c = -i - 3j + k are linearly dependent. If b is a unit vector, then the values of m and n could be:


m = −15m\ =\ -\frac{1}{\sqrt{5}} and n = 25n\ =\ \frac{2}{\sqrt{5}}

m = 15m\ =\ \frac{1}{\sqrt{5}} and n = 25n\ =\ \frac{2}{\sqrt{5}}

m = −15m\ =\ -\frac{1}{\sqrt{5}} and n = −25n\ =\ -\frac{2}{\sqrt{5}}

m = 13m\ =\ \frac{1}{\sqrt{3}} and n = −23n\ =\ -\frac{2}{\sqrt{3}}

m = 2 m\ =\ 2\ and n = 1n\ =\ 1

5.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

If the vectors a = mi -  m\sqrt{m}  j - 3k and b = mi +  m\sqrt{m}  j + 2k are perpendicular, then:

m = 0

m = 3 and m = -2

m = -3 and m = 2

m = 3

m = -2

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Let a = i + j, b = j + k, and c = 2i - j - 3k. Which of the following is false?

The vectors a and b have the same length

The angle between vectors a and b is  60\degree 

The vector a + c is parallel to the vector a - b

c = 2a - 3b

The vectors a, b, and c are linearly independent

7.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

A, B, and C are three points in space. To prove that ABC is a right-angled isosceles triangle, it is necessary to show:

∣AB‾∣∣AC‾∣ = AB‾ . AC‾\left|\overline{AB}\right|\left|\overline{AC}\right|\ =\ \overline{AB}\ .\ \overline{AC} and ∣AB‾∣ = ∣BC‾∣\left|\overline{AB}\right|\ =\ \left|\overline{BC}\right|

∣AB‾∣∣AC‾∣ = 2AB‾ . AC‾\left|\overline{AB}\right|\left|\overline{AC}\right|\ =\ \sqrt{2}\overline{AB}\ .\ \overline{AC} and ∣AB‾∣ = ∣BC‾∣\left|\overline{AB}\right|\ =\ \left|\overline{BC}\right|

∣AB‾∣ = ∣BC‾∣\left|\overline{AB}\right|\ =\ \left|\overline{BC}\right| and AB‾ . AC‾ = 0\overline{AB}\ .\ \overline{AC}\ =\ 0

AB‾+BC‾+CA‾ = 0→\overline{AB}+\overline{BC}+\overline{CA}\ =\ \overrightarrow{0} and BA‾ . BC‾ = 0\overline{BA}\ .\ \overline{BC}\ =\ 0

AB‾ + BC‾ + CA‾ = 0→\overline{AB}\ +\ \overline{BC}\ +\ \overline{CA}\ =\ \overrightarrow{0} and ∣AB‾∣ = ∣BC‾∣\left|\overline{AB}\right|\ =\ \left|\overline{BC}\right|

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