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 = 2 x + 2 j − k A = 2x + 2j - k A = 2 x + 2 j − k B = 6 i − 3 j + 2 k B = 6i - 3j + 2k B = 6 i − 3 j + 2 k
∣ A ∣ = 2 2 + 2 2 + ( − 1 ) 2 = 3 , ∣ B ∣ = 6 2 + ( − 3 ) 2 + 2 2 = 7. |A| = \sqrt{2^2 + 2^2 + (-1)^2} = 3, |B| = \sqrt{6^2 + (-3)^2 + 2^2} = 7. ∣ A ∣ = 2 2 + 2 2 + ( − 1 ) 2 = 3 , ∣ B ∣ = 6 2 + ( − 3 ) 2 + 2 2 = 7. ( A , B ) = 2 ⋅ 6 + 2 ⋅ ( − 3 ) + ( − 1 ) ⋅ 2 = 4. (A, B) = 2 \cdot 6 + 2 \cdot (-3) + (-1) \cdot 2 = 4. ( A , B ) = 2 ⋅ 6 + 2 ⋅ ( − 3 ) + ( − 1 ) ⋅ 2 = 4. cos ϕ = ( A , B ) ∣ A ∣ ⋅ ∣ B ∣ = 4 21 . \cos \phi = \frac{(A,B)}{|A| \cdot |B|} = \frac{4}{21}. cos ϕ = ∣ A ∣ ⋅ ∣ B ∣ ( A , B ) = 21 4 . ϕ = arccos 4 21 ≈ 79 ∘ . \phi = \arccos \frac{4}{21} \approx 79{}^\circ. ϕ = arccos 21 4 ≈ 79 ∘ .
Answer: a r c c o s 4 21 ≈ 79 ∘ arccos \frac{4}{21} \approx 79{}^\circ a rccos 21 4 ≈ 79 ∘
Question
2. Determine the value of a a a
A=2i+aj+k
and
B=4i-2j-2k
are perpendicular
a=5
a=3
a=1
a=7
Solution
If A = 2 i + a j + k A = 2i + aj + k A = 2 i + aj + k B = 4 i − 2 j − 2 k B = 4i - 2j - 2k B = 4 i − 2 j − 2 k
0 = ( A , B ) = 2 ⋅ 4 + a ⋅ ( − 2 ) + 1 ⋅ ( − 2 ) = 8 − 2 a − 2 = 6 − 2 a ⇒ 6 − 2 a = 0 ⇒ ⇒ 3 − a = 0 ⇒ a = 3. \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} 0 = ( A , B ) = 2 ⋅ 4 + a ⋅ ( − 2 ) + 1 ⋅ ( − 2 ) = 8 − 2 a − 2 = 6 − 2 a ⇒ 6 − 2 a = 0 ⇒ ⇒ 3 − a = 0 ⇒ a = 3.
Answer: a = 3 a = 3 a = 3
Question
3. Determine a unit vector perpendicular to the plane of
A=2i-6j-3k
and
B = 4 i + 3 j − k B = 4 i + 3 j - k B = 4 i + 3 j − k ± ( 37 i − 27 j + 67 k ) \pm (37i - 27j + 67k) ± ( 37 i − 27 j + 67 k )
± ( 35 i + 25 j − 65 k ) \pm (35i + 25j - 65k) ± ( 35 i + 25 j − 65 k )
± ( 14 i − 34 j − 12 k ) \pm (14i - 34j - 12k) ± ( 14 i − 34 j − 12 k )
± ( − 23 i − 13 j + 34 k ) \pm (-23i - 13j + 34k) ± ( − 23 i − 13 j + 34 k )
Solution
The equation of the plane is
∣ x 2 4 y − 6 3 z − 3 − 1 ∣ + D = 6 x + 6 z − 12 y + 24 z + 9 x + 2 y + D = 15 x − 10 y + 30 z + D . \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. ∣ ∣ x y z 2 − 6 − 3 4 3 − 1 ∣ ∣ + D = 6 x + 6 z − 12 y + 24 z + 9 x + 2 y + D = 15 x − 10 y + 30 z + D . n = ( 15 , − 10 , 30 ) n = (15, -10, 30) n = ( 15 , − 10 , 30 )
∣ n ∣ = 1 5 2 + ( − 10 ) 2 + 3 0 2 = 225 + 100 + 900 = 1225 = 35. |n| = \sqrt{15^2 + (-10)^2 + 30^2} = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35. ∣ n ∣ = 1 5 2 + ( − 10 ) 2 + 3 0 2 = 225 + 100 + 900 = 1225 = 35.
A unit vector perpendicular to the plane is
n 1 = n ∣ n ∣ = ( 15 35 , − 10 35 , 30 35 ) = ( 3 7 , − 2 7 , 6 7 ) , 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), n 1 = ∣ n ∣ n = ( 35 15 , − 35 10 , 35 30 ) = ( 7 3 , − 7 2 , 7 6 ) ,
hence n 1 = 3 7 i − 2 7 j + 6 7 k n_1 = \frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k n 1 = 7 3 i − 7 2 j + 7 6 k
Answer: 3 7 i − 2 7 j + 6 7 k \frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k 7 3 i − 7 2 j + 7 6 k
Question
4. Find the work done in moving an object along a vector r = 3 i + 2 j − 5 k r = 3i + 2j - 5k r = 3 i + 2 j − 5 k
3
5
7
9
Solution
The vector F ˉ = F 1 i + F 2 j + F 3 k \bar{F} = F_1i + F_2j + F_3k F ˉ = F 1 i + F 2 j + F 3 k
Then work done in moving an object along a vector r = 3 i + 2 j − 5 k r = 3i + 2j - 5k r = 3 i + 2 j − 5 k
A = ( F ˉ , r ˉ ) = 3 F 1 + 2 F 2 − 5 F 3 . A = (\bar{F}, \bar{r}) = 3F_1 + 2F_2 - 5F_3. A = ( F ˉ , r ˉ ) = 3 F 1 + 2 F 2 − 5 F 3 . Question
5. Given that
A = 2 i − j + 3 k A = 2i - j + 3k A = 2 i − j + 3 k
and
B = 3 i + 2 j − k , B = 3i + 2j - k, B = 3 i + 2 j − k ,
find
A·B
3
6
1
9
Solution
Given that A = 2 i − j + 3 k A = 2i - j + 3k A = 2 i − j + 3 k B = 3 i + 2 j − k B = 3i + 2j - k B = 3 i + 2 j − k
A ⋅ B = 2 ⋅ 3 + ( − 1 ) ⋅ 2 + 3 ⋅ ( − 1 ) = 1. A \cdot B = 2 \cdot 3 + (-1) \cdot 2 + 3 \cdot (-1) = 1. A ⋅ B = 2 ⋅ 3 + ( − 1 ) ⋅ 2 + 3 ⋅ ( − 1 ) = 1.
Answer: A ⋅ B = 1 A \cdot B = 1 A ⋅ B = 1
Question
6. If
A = 2 i − 3 j − k A = 2i - 3j - k A = 2 i − 3 j − k
and
B = i + 4 j − 2 k , B = i + 4j - 2k, B = i + 4 j − 2 k ,
find
( A + B ) × ( A − B ) (A + B) \times (A - B) ( A + B ) × ( A − B ) 3 i + 4 j + 25 k 2 i + 6 j + 2 k − 20 i − 6 j − 22 k − 3 i − 5 j − 25 k \begin{array}{l}
3i + 4j + 25k \\
2i + 6j + 2k \\
-20i - 6j - 22k \\
-3i - 5j - 25k \\
\end{array} 3 i + 4 j + 25 k 2 i + 6 j + 2 k − 20 i − 6 j − 22 k − 3 i − 5 j − 25 k Solution
If A = 2 i − 3 j − k A = 2i - 3j - k A = 2 i − 3 j − k B = i + 4 j − 2 k B = i + 4j - 2k B = i + 4 j − 2 k
A + B = ( 3 , 1 , − 3 ) , A − B = ( 1 , − 7 , 1 ) . A + B = (3, 1, -3), \quad A - B = (1, -7, 1). A + B = ( 3 , 1 , − 3 ) , A − B = ( 1 , − 7 , 1 ) . ( A + B ) × ( A − B ) = ∣ i j k 3 1 − 3 1 − 7 1 ∣ = ∣ 1 − 3 − 7 1 ∣ i − ∣ 3 − 3 1 1 ∣ j + ∣ 3 1 1 − 7 ∣ k = (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 = ( A + B ) × ( A − B ) = ∣ ∣ i 3 1 j 1 − 7 k − 3 1 ∣ ∣ = ∣ ∣ 1 − 7 − 3 1 ∣ ∣ i − ∣ ∣ 3 1 − 3 1 ∣ ∣ j + ∣ ∣ 3 1 1 − 7 ∣ ∣ k = ( 1 ⋅ 1 − ( − 7 ) ⋅ ( − 3 ) ) − ( 3 ⋅ 1 − 1 ⋅ ( − 3 ) ) j + ( 3 ⋅ ( − 7 ) − 1 ⋅ 1 ) k = − 20 i − 6 j − 22 k . (1 \cdot 1 - (-7) \cdot (-3)) - (3 \cdot 1 - 1 \cdot (-3))j + (3 \cdot (-7) - 1 \cdot 1)k = -20i - 6j - 22k. ( 1 ⋅ 1 − ( − 7 ) ⋅ ( − 3 )) − ( 3 ⋅ 1 − 1 ⋅ ( − 3 )) j + ( 3 ⋅ ( − 7 ) − 1 ⋅ 1 ) k = − 20 i − 6 j − 22 k .
Answer: ( A + B ) × ( A − B ) = − 20 i − 6 j − 22 k (A + B) \times (A - B) = -20i - 6j - 22k ( A + B ) × ( A − B ) = − 20 i − 6 j − 22 k
Question
7. If
A = 3 i − j + 2 k , A = 3i - j + 2k, A = 3 i − j + 2 k , B = 2 i + j − k , B = 2i + j - k, B = 2 i + j − k ,
and
C = i − 2 j + 2 k , C = i - 2j + 2k, C = i − 2 j + 2 k ,
find
( A × B ) × C (A \times B) \times C ( A × B ) × C
15i+15j-5k
5i+5j-5k
-10i+10j-5k
15i+10j-5k
Solution
If A = 3 i − j + 2 k A = 3i - j + 2k A = 3 i − j + 2 k B = 2 i + j − k B = 2i + j - k B = 2 i + j − k C = i − 2 j + 2 k C = i - 2j + 2k C = i − 2 j + 2 k
A × B = ∣ i j k 3 − 1 2 2 1 − 1 ∣ = i ∣ − 1 2 1 − 1 ∣ − j ∣ 3 2 2 − 1 ∣ + k ∣ 3 − 1 2 1 ∣ = ( ( − 1 ) ⋅ ( − 1 ) − 1 ⋅ 2 ) i − − j ( 3 ⋅ ( − 1 ) − 2 ⋅ 2 ) + k ( 3 ⋅ 1 − 2 ⋅ ( − 1 ) ) = − i + 7 j + 5 k . \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} A × B = ∣ ∣ i 3 2 j − 1 1 k 2 − 1 ∣ ∣ = i ∣ ∣ − 1 1 2 − 1 ∣ ∣ − j ∣ ∣ 3 2 2 − 1 ∣ ∣ + k ∣ ∣ 3 2 − 1 1 ∣ ∣ = (( − 1 ) ⋅ ( − 1 ) − 1 ⋅ 2 ) i − − j ( 3 ⋅ ( − 1 ) − 2 ⋅ 2 ) + k ( 3 ⋅ 1 − 2 ⋅ ( − 1 )) = − i + 7 j + 5 k . ( A × B ) × C = ∣ i j k − 1 7 5 1 − 2 2 ∣ = i ∣ 7 5 − 2 2 ∣ − j ∣ − 1 5 1 2 ∣ + k ∣ − 1 7 1 − 2 ∣ = ( 7 ⋅ 2 − ( − 2 ) ⋅ 5 ) i − − ( ( − 1 ) ⋅ 2 − 1 ⋅ 5 ) j + ( ( − 1 ) ⋅ ( − 2 ) − 1 ⋅ 7 ) k = 24 i + 7 j − 5 k . \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} ( A × B ) × C = ∣ ∣ i − 1 1 j 7 − 2 k 5 2 ∣ ∣ = i ∣ ∣ 7 − 2 5 2 ∣ ∣ − j ∣ ∣ − 1 1 5 2 ∣ ∣ + k ∣ ∣ − 1 1 7 − 2 ∣ ∣ = ( 7 ⋅ 2 − ( − 2 ) ⋅ 5 ) i − − (( − 1 ) ⋅ 2 − 1 ⋅ 5 ) j + (( − 1 ) ⋅ ( − 2 ) − 1 ⋅ 7 ) k = 24 i + 7 j − 5 k .
Answer: ( A × B ) × C = 24 i + 7 j − 5 k (A \times B) \times C = 24i + 7j - 5k ( A × B ) × C = 24 i + 7 j − 5 k
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
∣ x 2 4 y − 6 3 z − 3 − 1 ∣ + D = 6 x + 6 z − 12 y + 24 z + 9 x + 2 y + D = 15 x − 10 y + 30 z + D . \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. ∣ ∣ x y z 2 − 6 − 3 4 3 − 1 ∣ ∣ + D = 6 x + 6 z − 12 y + 24 z + 9 x + 2 y + D = 15 x − 10 y + 30 z + D .
A vector perpendicular to the plane is n = ( 15 , − 10 , 30 ) n = (15, -10, 30) n = ( 15 , − 10 , 30 )
∣ n ∣ = 225 + 100 + 900 = 1225 = 35. |n| = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35. ∣ n ∣ = 225 + 100 + 900 = 1225 = 35.
A unit vector perpendicular to the plane is
n 1 = n ∣ n ∣ = ( 15 35 , − − 10 35 , 30 35 ) = ( 3 7 , − 2 7 , 6 7 ) , 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), n 1 = ∣ n ∣ n = ( 35 15 , − 35 − 10 , 35 30 ) = ( 7 3 , − 7 2 , 7 6 ) ,
hence
n 1 = 3 7 i − 2 7 j + 6 7 k . n_1 = \frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k. n 1 = 7 3 i − 7 2 j + 7 6 k .
Answer: n 1 = 3 7 i − 2 7 j + 6 7 k . n_1 = \frac{3}{7}i - \frac{2}{7}j + \frac{6}{7}k. n 1 = 7 3 i − 7 2 j + 7 6 k .
Question
9. Evaluate
( 2 i − 3 j ) ⋅ [ ( i + j − k ) × ( 3 i − k ) ] (2i - 3j) \cdot [(i + j - k) \times (3i - k)] ( 2 i − 3 j ) ⋅ [( i + j − k ) × ( 3 i − k )]
www.AssignmentExpert.com
5
6
8
Solution
( i + j − k ) × ( 3 i − k ) = ∣ i j k 1 1 − 1 3 0 − 1 ∣ = − i − 2 j − 3 k . (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. ( i + j − k ) × ( 3 i − k ) = ∣ ∣ i 1 3 j 1 0 k − 1 − 1 ∣ ∣ = − i − 2 j − 3 k . ( 2 i − 3 j ) ⋅ [ ( i + j − k ) × ( 3 i − k ) ] = ( 2 i − 3 j ) ⋅ ( − i − 2 j − 3 k ) = 2 ⋅ ( − 1 ) + ( − 3 ) ⋅ ( − 2 ) + 0 ⋅ ( − 3 ) = − 2 + 6 + 0 = 4. (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. ( 2 i − 3 j ) ⋅ [( i + j − k ) × ( 3 i − k )] = ( 2 i − 3 j ) ⋅ ( − i − 2 j − 3 k ) = 2 ⋅ ( − 1 ) + ( − 3 ) ⋅ ( − 2 ) + 0 ⋅ ( − 3 ) = − 2 + 6 + 0 = 4.
Answer: 4.
Question
10. If
A = i − 2 j − 3 k , A = i - 2j - 3k, A = i − 2 j − 3 k , B = 2 i + j − k B = 2i + j - k B = 2 i + j − k
and
C = i + 3 j − 2 k , C = i + 3j - 2k, C = i + 3 j − 2 k ,
evaluate
( A × B ) ⋅ C (A \times B) \cdot C ( A × B ) ⋅ C
-25
11
15
-20
Solution
A × B = ∣ i j k 1 − 2 − 3 2 1 − 1 ∣ = ∣ − 2 − 3 1 − 1 ∣ i − ∣ 1 − 3 2 − 1 ∣ j + ∣ 1 − 2 2 1 ∣ k = ( ( − 2 ) ⋅ ( − 1 ) − 1 ⋅ ( − 3 ) ) i − 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 - A × B = ∣ ∣ i 1 2 j − 2 1 k − 3 − 1 ∣ ∣ = ∣ ∣ − 2 1 − 3 − 1 ∣ ∣ i − ∣ ∣ 1 2 − 3 − 1 ∣ ∣ j + ∣ ∣ 1 2 − 2 1 ∣ ∣ k = (( − 2 ) ⋅ ( − 1 ) − 1 ⋅ ( − 3 )) i − ( 1 ⋅ ( − 1 ) − 2 ⋅ ( − 3 ) ) j + ( 1 ⋅ 1 − 2 ⋅ ( − 2 ) ) k = 5 i − 5 j + 5 k . (1 \cdot (-1) - 2 \cdot (-3)) j + (1 \cdot 1 - 2 \cdot (-2)) k = 5i - 5j + 5k. ( 1 ⋅ ( − 1 ) − 2 ⋅ ( − 3 )) j + ( 1 ⋅ 1 − 2 ⋅ ( − 2 )) k = 5 i − 5 j + 5 k . ( A × B ) ⋅ C = ( 5 i − 5 j + 5 k ) ⋅ ( i + 3 j − 2 k ) = 5 ⋅ 1 + ( − 5 ) ⋅ 3 + 5 ⋅ ( − 2 ) = 5 − 15 − 10 = − 20. (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. ( A × B ) ⋅ C = ( 5 i − 5 j + 5 k ) ⋅ ( i + 3 j − 2 k ) = 5 ⋅ 1 + ( − 5 ) ⋅ 3 + 5 ⋅ ( − 2 ) = 5 − 15 − 10 = − 20.
Answer: ( A × B ) ⋅ C = − 20 (A \times B) \cdot C = -20 ( A × B ) ⋅ C = − 20
#339914 on Dec 2023
Comments