Question

1 Find the angle between
A=2x+2j−k
and
B=6i−3j+2k
600
450
690
790

2 Determine the value of a so that
A=2i+aj+k
and
B=4i−2j−2k
are perpendicular
a=5
a=3
a=1
a=7

3 Determine a unit vector perpendicular to the plane of
A=2i−6j−3k
and
B=4i+3j−k
±(37i−27j+67k)
±(35i+25j−65k)
±(14i−34j−12k)
±(−23i−13j+34k)

4 Find the work done in moving an object along a vector
r=3i+2j−5k
3
5
7
9

5 Given that
A=2i−j+3k
and
B=3i+2j−k
, find
A⋅B
3
6
1
9

6 If
A=2i−3j−k
and
B=i+4j−2k
, find
(A+B+×(A−B)
3i+4j+25k
2i+6j+2k
−20i−6j−22k
−3i−5j−25k

7 If
A=3i−j+2k
,
B=2i+j−k
and
C=i−2j+2k
, find
(A×B)×C
15i+15j−5k
5i+5j−5k
−10i+10j−5k
15i+10j−5k

8 Determine a unit vector perpendicular to the plane of
A=2i−6j−3k
and
B=4i+3j−k
35i−25j+65
17i−37j+47
37i−27j+67
27i−47j+57

9 Evaluate
(2i−3j)⋅[(i+j−k)×(3i−k)]
4
5
6
8

10 If
A=i−2j−3k
,
B=2i+j−k
and
C=i+3j−2k
,evaluate
(A×B)⋅C
-25
11
15
-20

Answer 1

Answer on Question #58423 – Math – Vector Calculus

Question

1. Find the angle between

A=2x+2j-k

and

B=6i-3j+2k

600

450

690

790

Solution

If $A = 2x + 2j - k$ and $B = 6i - 3j + 2k$, then

$$|A| = \sqrt{2^2 + 2^2 + (-1)^2} = 3, |B| = \sqrt{6^2 + (-3)^2 + 2^2} = 7.$$ $$(A, B) = 2 \cdot 6 + 2 \cdot (-3) + (-1) \cdot 2 = 4.$$ $$\cos \phi = \frac{(A,B)}{|A| \cdot |B|} = \frac{4}{21}.$$ $$\phi = \arccos \frac{4}{21} \approx 79{}^\circ.$$

Answer: $arccos \frac{4}{21} \approx 79{}^\circ$.

Question

2. Determine the value of $a$ so that

A=2i+aj+k

and

B=4i-2j-2k

are perpendicular

a=5

a=3

a=1

a=7

Solution

If $A = 2i + aj + k$ and $B = 4i - 2j - 2k$ are perpendicular, then

$$\begin{array}{l} 0 = (A, B) = 2 \cdot 4 + a \cdot (-2) + 1 \cdot (-2) = 8 - 2a - 2 = 6 - 2a \Rightarrow 6 - 2a = 0 \Rightarrow \\ \Rightarrow 3 - a = 0 \Rightarrow a = 3. \end{array}$$

Answer: $a = 3$.

Question

3. Determine a unit vector perpendicular to the plane of

A=2i-6j-3k

and

$$B = 4 i + 3 j - k$$

$\pm (37i - 27j + 67k)$

$\pm (35i + 25j - 65k)$

$\pm (14i - 34j - 12k)$

$\pm (-23i - 13j + 34k)$

Solution

The equation of the plane is

$$\left| \begin{array}{ccc} x & 2 & 4 \\ y & -6 & 3 \\ z & -3 & -1 \end{array} \right| + D = 6x + 6z - 12y + 24z + 9x + 2y + D = 15x - 10y + 30z + D.$$

$n = (15, -10, 30)$ is a vector perpendicular to the plane,

$$|n| = \sqrt{15^2 + (-10)^2 + 30^2} = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35.$$

A unit vector perpendicular to the plane is

$$n_1 = \frac{n}{|n|} = \left(\frac{15}{35}, -\frac{10}{35}, \frac{30}{35}\right) = \left(\frac{3}{7}, -\frac{2}{7}, \frac{6}{7}\right),$$

hence $n_1 = \frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k$.

Answer: $\frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k$.

Question

4. Find the work done in moving an object along a vector $r = 3i + 2j - 5k$

3

5

7

9

Solution

The vector $\bar{F} = F_1i + F_2j + F_3k$ is needed.

Then work done in moving an object along a vector $r = 3i + 2j - 5k$ is

$$A = (\bar{F}, \bar{r}) = 3F_1 + 2F_2 - 5F_3.$$

Question

5. Given that

$$A = 2i - j + 3k$$

and

$$B = 3i + 2j - k,$$

find

A·B

3

6

1

9

Solution

Given that $A = 2i - j + 3k$ and $B = 3i + 2j - k$,

$$A \cdot B = 2 \cdot 3 + (-1) \cdot 2 + 3 \cdot (-1) = 1.$$

Answer: $A \cdot B = 1$.

Question

6. If

$$A = 2i - 3j - k$$

and

$$B = i + 4j - 2k,$$

find

$$(A + B) \times (A - B)$$ $$\begin{array}{l} 3i + 4j + 25k \\ 2i + 6j + 2k \\ -20i - 6j - 22k \\ -3i - 5j - 25k \\ \end{array}$$

Solution

If $A = 2i - 3j - k$ and $B = i + 4j - 2k$,

$$A + B = (3, 1, -3), \quad A - B = (1, -7, 1).$$ $$(A + B) \times (A - B) = \begin{vmatrix} i & j & k \\ 3 & 1 & -3 \\ 1 & -7 & 1 \end{vmatrix} = \begin{vmatrix} 1 & -3 \\ -7 & 1 \end{vmatrix} i - \begin{vmatrix} 3 & -3 \\ 1 & 1 \end{vmatrix} j + \begin{vmatrix} 3 & 1 \\ 1 & -7 \end{vmatrix} k =$$ $$(1 \cdot 1 - (-7) \cdot (-3)) - (3 \cdot 1 - 1 \cdot (-3))j + (3 \cdot (-7) - 1 \cdot 1)k = -20i - 6j - 22k.$$

Answer: $(A + B) \times (A - B) = -20i - 6j - 22k$.

Question

7. If

$$A = 3i - j + 2k,$$ $$B = 2i + j - k,$$

and

$$C = i - 2j + 2k,$$

find

$$(A \times B) \times C$$

15i+15j-5k

5i+5j-5k

-10i+10j-5k

15i+10j-5k

Solution

If $A = 3i - j + 2k$, $B = 2i + j - k$, and $C = i - 2j + 2k$,

$$\begin{array}{l} A \times B = \left| \begin{array}{ccc} i & j & k \\ 3 & -1 & 2 \\ 2 & 1 & -1 \end{array} \right| = i \left| \begin{array}{cc} -1 & 2 \\ 1 & -1 \end{array} \right| - j \left| \begin{array}{cc} 3 & 2 \\ 2 & -1 \end{array} \right| + k \left| \begin{array}{cc} 3 & -1 \\ 2 & 1 \end{array} \right| = ((-1) \cdot (-1) - 1 \cdot 2)i - \\ - j(3 \cdot (-1) - 2 \cdot 2) + k(3 \cdot 1 - 2 \cdot (-1)) = -i + 7j + 5k. \end{array}$$ $$\begin{array}{l} (A \times B) \times C = \left| \begin{array}{ccc} i & j & k \\ -1 & 7 & 5 \\ 1 & -2 & 2 \end{array} \right| = i \left| \begin{array}{cc} 7 & 5 \\ -2 & 2 \end{array} \right| - j \left| \begin{array}{cc} -1 & 5 \\ 1 & 2 \end{array} \right| + k \left| \begin{array}{cc} -1 & 7 \\ 1 & -2 \end{array} \right| = (7 \cdot 2 - (-2) \cdot 5)i - \\ - ((-1) \cdot 2 - 1 \cdot 5)j + ((-1) \cdot (-2) - 1 \cdot 7)k = 24i + 7j - 5k. \end{array}$$

Answer: $(A \times B) \times C = 24i + 7j - 5k$.

Question

8. Determine a unit vector perpendicular to the plane of

A=2i-6j-3k

and

B=4i+3j-k

35i-25j+65

17i-37j+47

37i-27j+67

27i-47j+57

Solution

The equation of the plane is

$$\left| \begin{array}{ccc} x & 2 & 4 \\ y & -6 & 3 \\ z & -3 & -1 \end{array} \right| + D = 6x + 6z - 12y + 24z + 9x + 2y + D = 15x - 10y + 30z + D.$$

A vector perpendicular to the plane is $n = (15, -10, 30)$,

$$|n| = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35.$$

A unit vector perpendicular to the plane is

$$n_1 = \frac{n}{|n|} = \left(\frac{15}{35}, -\frac{-10}{35}, \frac{30}{35}\right) = \left(\frac{3}{7}, -\frac{2}{7}, \frac{6}{7}\right),$$

hence

$$n_1 = \frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k.$$

Answer: $n_1 = \frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k.$

Question

9. Evaluate

$$(2i - 3j) \cdot [(i + j - k) \times (3i - k)]$$

www.AssignmentExpert.com

5

6

8

Solution

$$(i + j - k) \times (3i - k) = \left| \begin{array}{ccc} i & j & k \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{array} \right| = -i - 2j - 3k.$$ $$(2i - 3j) \cdot [(i + j - k) \times (3i - k)] = (2i - 3j) \cdot (-i - 2j - 3k) = 2 \cdot (-1) + (-3) \cdot (-2) + 0 \cdot (-3) = -2 + 6 + 0 = 4.$$

Answer: 4.

Question

10. If

$$A = i - 2j - 3k,$$ $$B = 2i + j - k$$

and

$$C = i + 3j - 2k,$$

evaluate

$$(A \times B) \cdot C$$

-25

11

15

-20

Solution

$$A \times B = \left| \begin{array}{ccc} i & j & k \\ 1 & -2 & -3 \\ 2 & 1 & -1 \end{array} \right| = \left| \begin{array}{cc} -2 & -3 \\ 1 & -1 \end{array} \right| i - \left| \begin{array}{cc} 1 & -3 \\ 2 & -1 \end{array} \right| j + \left| \begin{array}{cc} 1 & -2 \\ 2 & 1 \end{array} \right| k = ((-2) \cdot (-1) - 1 \cdot (-3)) i -$$ $$(1 \cdot (-1) - 2 \cdot (-3)) j + (1 \cdot 1 - 2 \cdot (-2)) k = 5i - 5j + 5k.$$ $$(A \times B) \cdot C = (5i - 5j + 5k) \cdot (i + 3j - 2k) = 5 \cdot 1 + (-5) \cdot 3 + 5 \cdot (-2) = 5 - 15 - 10 = -20.$$

Answer: $(A \times B) \cdot C = -20$.