Answer on Question #52239 – Math – Vector Calculus
6. If U = I + 3 j − 2 k U = I + 3j - 2k U = I + 3 j − 2 k V = 4 i − 2 j − 4 k V = 4i - 2j - 4k V = 4 i − 2 j − 4 k ( 2 U + V ) ⋅ ( U − 2 V ) (2U + V) \cdot (U - 2V) ( 2 U + V ) ⋅ ( U − 2 V )
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7. If r 4 = r 1 + r 2 + r 3 r4 = r1 + r2 + r3 r 4 = r 1 + r 2 + r 3
Which of the vectors are linearly dependence on each r 1 r1 r 1
r1
r3
r2
r4
8. A vector quantity has both magnitude and ---.
direction
time
magnitude
scalar
9. What does the symbol i , j , k i, j, k i , j , k
distance
velocity
unit vector
angle
10 Evaluate the vectors ( 2 i − 3 j ) (2i - 3j) ( 2 i − 3 j ) [ ( I + j − k ) × ( 3 i − k ) ] [(I + j - k) \times (3i - k)] [( I + j − k ) × ( 3 i − k )]
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Solution
6.
**Method 1**
( 2 U + V ) ⋅ ( U − 2 V ) = = [ 2 ( i + 3 j − 2 k ) + ( 4 i − 2 j − 4 k ) ] ⋅ [ ( i + 3 j − 2 k ) − 2 ( 4 i − 2 j − 4 k ) ] = = ( 6 i + 4 j − 8 k ) ⋅ ( − 7 i + 7 j + 6 k ) = − 42 + 28 − 48 = − 62. \begin{array}{l}
(2U + V) \cdot (U - 2V) = \\
= \left[ 2(i + 3j - 2k) + (4i - 2j - 4k) \right] \cdot \left[ (i + 3j - 2k) - 2(4i - 2j - 4k) \right] = \\
= (6i + 4j - 8k) \cdot (-7i + 7j + 6k) = -42 + 28 - 48 = -62.
\end{array} ( 2 U + V ) ⋅ ( U − 2 V ) = = [ 2 ( i + 3 j − 2 k ) + ( 4 i − 2 j − 4 k ) ] ⋅ [ ( i + 3 j − 2 k ) − 2 ( 4 i − 2 j − 4 k ) ] = = ( 6 i + 4 j − 8 k ) ⋅ ( − 7 i + 7 j + 6 k ) = − 42 + 28 − 48 = − 62.
**Method 2**
( 2 U + V ) ⋅ ( U − 2 V ) = 2 U ⋅ U − 2 U ⋅ 2 V + V ⋅ U − V ⋅ 2 V = = 2 U 2 − 3 U V − 2 V 2 = 2 ( 1 2 + 3 2 + ( − 2 ) 2 ) − 3 ( 1 ⋅ 4 + 3 ⋅ ( − 2 ) + ( − 2 ) ⋅ ( − 4 ) ) − 2 ( 4 2 + ( − 2 ) 2 + ( − 4 ) 2 ) = 2 ⋅ 14 − 3 ⋅ 6 − 2 ⋅ 36 = 28 − 18 − 72 = − 62. \begin{array}{l}
(2U + V) \cdot (U - 2V) = 2U \cdot U - 2U \cdot 2V + V \cdot U - V \cdot 2V = \\
= 2U^2 - 3UV - 2V^2 = 2(1^2 + 3^2 + (-2)^2) - 3(1 \cdot 4 + 3 \cdot (-2) + (-2) \cdot (-4)) - 2(4^2 + (-2)^2 + (-4)^2) = 2 \cdot 14 - 3 \cdot 6 - 2 \cdot 36 = 28 - 18 - 72 = -62.
\end{array} ( 2 U + V ) ⋅ ( U − 2 V ) = 2 U ⋅ U − 2 U ⋅ 2 V + V ⋅ U − V ⋅ 2 V = = 2 U 2 − 3 U V − 2 V 2 = 2 ( 1 2 + 3 2 + ( − 2 ) 2 ) − 3 ( 1 ⋅ 4 + 3 ⋅ ( − 2 ) + ( − 2 ) ⋅ ( − 4 )) − 2 ( 4 2 + ( − 2 ) 2 + ( − 4 ) 2 ) = 2 ⋅ 14 − 3 ⋅ 6 − 2 ⋅ 36 = 28 − 18 − 72 = − 62.
**Answer: -62.**
7. r 4 r_4 r 4
8. direction
9. unit vector
10.
Method 1.
If a = ( a 1 ; a 2 ; a 3 ) a = (a_1; a_2; a_3) a = ( a 1 ; a 2 ; a 3 ) b = ( b 1 ; b 2 ; b 3 ) b = (b_1; b_2; b_3) b = ( b 1 ; b 2 ; b 3 ) c = ( c 1 ; c 2 ; c 3 ) c = (c_1; c_2; c_3) c = ( c 1 ; c 2 ; c 3 ) a ⋅ ( b × c ) a \cdot (b \times c) a ⋅ ( b × c ) a , b , c a, b, c a , b , c
a ⋅ ( b × c ) = ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ a \cdot (b \times c) = \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| a ⋅ ( b × c ) = ∣ ∣ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ∣ ∣
(in other words, it is the determinant).
Thus, ( 2 i − 3 j ) ⋅ [ ( i + j − k ) × ( 3 i − k ) ] = ∣ 2 − 3 0 1 1 − 1 3 0 − 1 ∣ = (2i - 3j) \cdot [(i + j - k) \times (3i - k)] = \left| \begin{array}{ccc} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{array} \right| = ( 2 i − 3 j ) ⋅ [( i + j − k ) × ( 3 i − k )] = ∣ ∣ 2 1 3 − 3 1 0 0 − 1 − 1 ∣ ∣ =
= ( subtract the third row from the second row ) = ∣ 2 − 3 0 − 2 1 0 3 0 − 1 ∣ = = ( expand the determinant over the third column ) = = − 1 ∣ 2 − 3 − 2 1 ∣ = = − ( 2 ⋅ 1 − ( − 2 ) ⋅ ( − 3 ) ) = − ( 2 − 6 ) = 4 \begin{array}{l}
= (\text{subtract the third row from the second row}) = \left| \begin{array}{ccc} 2 & -3 & 0 \\ -2 & 1 & 0 \\ 3 & 0 & -1 \end{array} \right| = \\
= (\text{expand the determinant over the third column}) = \\
= -1 \left| \begin{array}{cc} 2 & -3 \\ -2 & 1 \end{array} \right| = \\
= -\left(2 \cdot 1 - (-2) \cdot (-3)\right) = -(2 - 6) = 4 \\
\end{array} = ( subtract the third row from the second row ) = ∣ ∣ 2 − 2 3 − 3 1 0 0 0 − 1 ∣ ∣ = = ( expand the determinant over the third column ) = = − 1 ∣ ∣ 2 − 2 − 3 1 ∣ ∣ = = − ( 2 ⋅ 1 − ( − 2 ) ⋅ ( − 3 ) ) = − ( 2 − 6 ) = 4
To express ( i + j − k ) × ( 3 i − k ) (i + j - k) \times (3i - k) ( i + j − k ) × ( 3 i − k )
Method 2
( i + j − k ) × ( 3 i − k ) = ∣ i j k 1 1 − 1 3 0 − 1 ∣ = ∣ 1 − 1 0 − 1 ∣ i − ∣ 1 − 1 3 − 1 ∣ j + + ∣ 1 1 3 0 ∣ k = ( 1 ⋅ ( − 1 ) − 0 ⋅ ( − 1 ) ) i − ( 1 ⋅ ( − 1 ) − 3 ⋅ ( − 1 ) ) j + + ( 1 ⋅ 0 − 3 ⋅ 1 ) k = − 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| = \left| \begin{array}{cc} 1 & -1 \\ 0 & -1 \end{array} \right| i - \left| \begin{array}{cc} 1 & -1 \\ 3 & -1 \end{array} \right| j + \\
+ \left| \begin{array}{cc} 1 & 1 \\ 3 & 0 \end{array} \right| k = (1 \cdot (-1) - 0 \cdot (-1)) i - (1 \cdot (-1) - 3 \cdot (-1)) j + \\
+ (1 \cdot 0 - 3 \cdot 1) k = -i - 2j - 3k ( i + j − k ) × ( 3 i − k ) = ∣ ∣ i 1 3 j 1 0 k − 1 − 1 ∣ ∣ = ∣ ∣ 1 0 − 1 − 1 ∣ ∣ i − ∣ ∣ 1 3 − 1 − 1 ∣ ∣ j + + ∣ ∣ 1 3 1 0 ∣ ∣ k = ( 1 ⋅ ( − 1 ) − 0 ⋅ ( − 1 )) i − ( 1 ⋅ ( − 1 ) − 3 ⋅ ( − 1 )) j + + ( 1 ⋅ 0 − 3 ⋅ 1 ) k = − i − 2 j − 3 k Method 3
By properties of vector (or cross) product, i × i = j × j = k × k = 0 i \times i = j \times j = k \times k = 0 i × i = j × j = k × k = 0 i × j = k i \times j = k i × j = k j × k = i j \times k = i j × k = i k × i = j k \times i = j k × i = j
a × b = − b × a , a \times b = -b \times a, a × b = − b × a ,
hence, i × k = − j i \times k = -j i × k = − j j × i = − k j \times i = -k j × i = − k
Besides,
a × ( b + c ) = a × b + a × c , a \times (b + c) = a \times b + a \times c, a × ( b + c ) = a × b + a × c , ( λ a ) × b = a × ( λ b ) = λ ( a × b ) (\lambda a) \times b = a \times (\lambda b) = \lambda (a \times b) ( λa ) × b = a × ( λb ) = λ ( a × b )
These formulae allow to simplify
( i + j − k ) × ( 3 i − k ) = ( i × 3 i ) + ( i × ( − k ) ) + ( j × 3 i ) + ( j × ( − k ) ) + + ( − k × 3 i ) + ( − k × ( − k ) ) = 3 ( i × i ) − ( i × k ) + 3 ( j × i ) − ( j × k ) − − 3 ( k × i ) + ( k × k ) = 0 + j − 3 k − i − 3 j + 0 = − i − 2 j − 3 k (i + j - k) \times (3i - k) = (i \times 3i) + (i \times (-k)) + (j \times 3i) + (j \times (-k)) + \\
+ (-k \times 3i) + (-k \times (-k)) = 3(i \times i) - (i \times k) + 3(j \times i) - (j \times k) - \\
- 3(k \times i) + (k \times k) = 0 + j - 3k - i - 3j + 0 = -i - 2j - 3k ( i + j − k ) × ( 3 i − k ) = ( i × 3 i ) + ( i × ( − k )) + ( j × 3 i ) + ( j × ( − k )) + + ( − k × 3 i ) + ( − k × ( − k )) = 3 ( i × i ) − ( i × k ) + 3 ( j × i ) − ( j × k ) − − 3 ( k × i ) + ( k × k ) = 0 + j − 3 k − i − 3 j + 0 = − i − 2 j − 3 k
Next, by properties of scalar (or dot) product,
a ⋅ ( b + c ) = a ⋅ b + a ⋅ c , i ⋅ j = i ⋅ k = j ⋅ k = 0 , i ⋅ i = j ⋅ j = k ⋅ k = 1. a \cdot (b + c) = a \cdot b + a \cdot c, i \cdot j = i \cdot k = j \cdot k = 0, i \cdot i = j \cdot j = k \cdot k = 1. a ⋅ ( b + c ) = a ⋅ b + a ⋅ c , i ⋅ j = i ⋅ k = j ⋅ k = 0 , i ⋅ i = j ⋅ j = k ⋅ k = 1.
If vectors are given by their coordinates, a = ( a x , a y , a z ) a = (a_x, a_y, a_z) a = ( a x , a y , a z ) b = ( b x , b y , b z ) b = (b_x, b_y, b_z) b = ( b x , b y , b z )
a ⋅ b = ( a x i + a y j + a z k ) ( b x i + b y j + b z k ) = a x b x + a y b y + a z b z . a \cdot b = (a _ {x} i + a _ {y} j + a _ {z} k) (b _ {x} i + b _ {y} j + b _ {z} k) = a _ {x} b _ {x} + a _ {y} b _ {y} + a _ {z} b _ {z}. a ⋅ b = ( a x i + a y j + a z k ) ( b x i + b y j + b z k ) = a x b x + a y b y + a z b z .
These formulae allow to simplify
( 2 i − 3 j ) ⋅ [ ( i + j − k ) × ( 3 i − k ) ] = ( 2 i − 3 j ) ⋅ ( − i − 2 j − 3 k ) = = 2 ⋅ ( − 1 ) + ( − 3 ) ⋅ ( − 2 ) = − 2 + 6 = 4. \begin{array}{l} (2 i - 3 j) \cdot [ (i + j - k) \times (3 i - k) ] = (2 i - 3 j) \cdot (- i - 2 j - 3 k) = \\ = 2 \cdot (- 1) + (- 3) \cdot (- 2) = - 2 + 6 = 4. \\ \end{array} ( 2 i − 3 j ) ⋅ [( i + j − k ) × ( 3 i − k )] = ( 2 i − 3 j ) ⋅ ( − i − 2 j − 3 k ) = = 2 ⋅ ( − 1 ) + ( − 3 ) ⋅ ( − 2 ) = − 2 + 6 = 4.
Answer: 4.
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