Answer on Question #52237 – Math – Vector Calculus
What are vectors that are not parallel to the same line, called?
- scalar
- collinear vectors
- non-collinear vectors
- vectors
Answer: non-collinear vectors.
7 Find the vector product axb. If a = 2 i + 3 j + 4 k a = 2i + 3j + 4k a = 2 i + 3 j + 4 k b = 5 i − 2 j + k b = 5i - 2j + k b = 5 i − 2 j + k
11 i + 18 j − 19 k 11i + 18j - 19k 11 i + 18 j − 19 k 2 j + 3 k 2j + 3k 2 j + 3 k 5 i − 6 j + 7 k 5i - 6j + 7k 5 i − 6 j + 7 k 4 i − 6 j + 11 k 4i - 6j + 11k 4 i − 6 j + 11 k
Solution
a ⃗ x b ⃗ = ∣ i j k 2 3 4 5 − 2 1 ∣ = ∣ 3 4 − 2 1 ∣ i − ∣ 2 4 5 1 ∣ j + ∣ 2 3 5 − 2 ∣ k = = i ( 3 ⋅ 1 − 4 ( − 2 ) ) + j ( 4 ⋅ 5 − 1 ⋅ 2 ) + k ( 2 ⋅ ( − 2 ) − 3 ⋅ 5 ) = 11 i + 18 j − 19 k . \begin{array}{l}
\vec{a} x \vec{b} = \left| \begin{array}{ccc} i & j & k \\ 2 & 3 & 4 \\ 5 & -2 & 1 \end{array} \right| = \left| \begin{array}{ccc} 3 & 4 \\ -2 & 1 \end{array} \right| i - \left| \begin{array}{ccc} 2 & 4 \\ 5 & 1 \end{array} \right| j + \left| \begin{array}{ccc} 2 & 3 \\ 5 & -2 \end{array} \right| k = \\
= i(3 \cdot 1 - 4(-2)) + j(4 \cdot 5 - 1 \cdot 2) + k(2 \cdot (-2) - 3 \cdot 5) = 11i + 18j - 19k.
\end{array} a x b = ∣ ∣ i 2 5 j 3 − 2 k 4 1 ∣ ∣ = ∣ ∣ 3 − 2 4 1 ∣ ∣ i − ∣ ∣ 2 5 4 1 ∣ ∣ j + ∣ ∣ 2 5 3 − 2 ∣ ∣ k = = i ( 3 ⋅ 1 − 4 ( − 2 )) + j ( 4 ⋅ 5 − 1 ⋅ 2 ) + k ( 2 ⋅ ( − 2 ) − 3 ⋅ 5 ) = 11 i + 18 j − 19 k .
Answer: 11 i + 18 j − 19 k 11i + 18j - 19k 11 i + 18 j − 19 k
8 A ... line AB occurs when the point A is fixed.
- free vector
- position vector
- force
- null vector
Answer: position vector.
9 A north-easterly wind of 20 knots is a ... quantity.
- scalar
- weight
- vector
distance
Answer: vector.
10 Given that a = 5 i + 2 j − k a = 5i + 2j - k a = 5 i + 2 j − k b = l − 3 j + k b = l - 3j + k b = l − 3 j + k ( a + b ) × ( a − b ) (a + b) \times (a - b) ( a + b ) × ( a − b )
2 i − 12 j − 34 k 2i - 12j - 34k 2 i − 12 j − 34 k 2 i + 12 j + 34 k 2i + 12j + 34k 2 i + 12 j + 34 k 2 i − 3 j + 12 j 2i - 3j + 12j 2 i − 3 j + 12 j 2 i + 2 k 2i + 2k 2 i + 2 k Solution
Method 1 (straight-forward calculation)
( a ⃗ + b ⃗ ) = 5 i + 2 j − k + i − 3 j + k = 6 i − j . (\vec{a} + \vec{b}) = 5i + 2j - k + i - 3j + k = 6i - j. ( a + b ) = 5 i + 2 j − k + i − 3 j + k = 6 i − j . ( a ⃗ − b ⃗ ) = 5 i + 2 j − k − i + 3 j − k = 4 i + 5 j − 2 k . (\vec{a} - \vec{b}) = 5i + 2j - k - i + 3j - k = 4i + 5j - 2k. ( a − b ) = 5 i + 2 j − k − i + 3 j − k = 4 i + 5 j − 2 k . ( a ⃗ + b ⃗ ) × ( a ⃗ − b ⃗ ) = ∣ i j k 6 − 1 0 4 5 − 2 ∣ = ∣ − 1 0 5 − 2 ∣ i − ∣ 6 0 4 − 2 ∣ j + ∣ 6 − 1 4 5 ∣ k = ( − 1 ⋅ ( − 2 ) − 5 ⋅ 0 ) i − ( 6 ⋅ ( − 2 ) − 4 ⋅ 0 ) j + ( 6 ⋅ 5 − 4 ⋅ ( − 1 ) ) k = 2 i + 12 j + 34 k . \begin{array}{l}
(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} i & j & k \\ 6 & -1 & 0 \\ 4 & 5 & -2 \end{vmatrix} = \begin{vmatrix} -1 & 0 \\ 5 & -2 \end{vmatrix} i - \begin{vmatrix} 6 & 0 \\ 4 & -2 \end{vmatrix} j + \begin{vmatrix} 6 & -1 \\ 4 & 5 \end{vmatrix} k \\
= (-1 \cdot (-2) - 5 \cdot 0)i - (6 \cdot (-2) - 4 \cdot 0)j + (6 \cdot 5 - 4 \cdot (-1))k = 2i + 12j + 34k.
\end{array} ( a + b ) × ( a − b ) = ∣ ∣ i 6 4 j − 1 5 k 0 − 2 ∣ ∣ = ∣ ∣ − 1 5 0 − 2 ∣ ∣ i − ∣ ∣ 6 4 0 − 2 ∣ ∣ j + ∣ ∣ 6 4 − 1 5 ∣ ∣ k = ( − 1 ⋅ ( − 2 ) − 5 ⋅ 0 ) i − ( 6 ⋅ ( − 2 ) − 4 ⋅ 0 ) j + ( 6 ⋅ 5 − 4 ⋅ ( − 1 )) k = 2 i + 12 j + 34 k . Method 2 (application of cross product properties)
The following properties of the cross product will be used:
a ⃗ × b ⃗ = − b ⃗ × a ⃗ ; a ⃗ × ( b ⃗ + c ⃗ ) = a ⃗ × b ⃗ + a ⃗ × c ⃗ ; a ⃗ × a ⃗ = b ⃗ × b ⃗ = 0 ⃗ ; \vec{a} \times \vec{b} = -\vec{b} \times \vec{a}; \quad \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}; \quad \vec{a} \times \vec{a} = \vec{b} \times \vec{b} = \vec{0}; a × b = − b × a ; a × ( b + c ) = a × b + a × c ; a × a = b × b = 0 ; ( λ a ⃗ ) × b ⃗ = a ⃗ × ( λ b ⃗ ) = λ ( a ⃗ × b ⃗ ) (\lambda \vec{a}) \times \vec{b} = \vec{a} \times (\lambda \vec{b}) = \lambda (\vec{a} \times \vec{b}) ( λ a ) × b = a × ( λ b ) = λ ( a × b ) a ⃗ × b ⃗ = ∣ i j k a x a y a z b x b y b z ∣ = ∣ a y a z b y b z ∣ i − ∣ a x a z b x b z ∣ j + ∣ a x a y b x b y ∣ k = = ( a y b z − b y a z ) i + ( a z b x − a x b z ) j + ( a x b y − a y b x ) k \begin{array}{l}
\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} = \begin{vmatrix} a_y & a_z \\ b_y & b_z \end{vmatrix} i - \begin{vmatrix} a_x & a_z \\ b_x & b_z \end{vmatrix} j + \begin{vmatrix} a_x & a_y \\ b_x & b_y \end{vmatrix} k = \\
= (a_y b_z - b_y a_z)i + (a_z b_x - a_x b_z)j + (a_x b_y - a_y b_x)k
\end{array} a × b = ∣ ∣ i a x b x j a y b y k a z b z ∣ ∣ = ∣ ∣ a y b y a z b z ∣ ∣ i − ∣ ∣ a x b x a z b z ∣ ∣ j + ∣ ∣ a x b x a y b y ∣ ∣ k = = ( a y b z − b y a z ) i + ( a z b x − a x b z ) j + ( a x b y − a y b x ) k
Simplify
( a ⃗ + b ⃗ ) × ( a ⃗ − b ⃗ ) = ( a ⃗ × a ⃗ ) + ( a ⃗ × ( − b ⃗ ) ) + ( b ⃗ × a ⃗ ) + ( b ⃗ × ( − b ⃗ ) ) = ( a ⃗ × a ⃗ ) − ( a ⃗ × b ⃗ ) + ( b ⃗ × a ⃗ ) − ( b ⃗ × b ⃗ ) = 0 ⃗ − ( a ⃗ × b ⃗ ) − ( a ⃗ × b ⃗ ) − 0 ⃗ = − 2 ( a ⃗ × b ⃗ ) = − 2 ∣ i j k 5 2 − 1 1 − 3 1 ∣ = − 2 ( 2 − 1 − 3 1 ) i − ∣ 5 − 1 1 1 ∣ j + ∣ 5 2 1 − 3 ∣ k = − 2 ( ( 2 ⋅ 1 − ( − 3 ) ⋅ ( − 1 ) ) i − ( 5 ⋅ 1 − 1 ⋅ ( − 1 ) ) j + ( 5 ⋅ ( − 3 ) − 1 ⋅ 2 ) k ) = − 2 ( − i − 6 j − 17 k ) = 2 i + 12 j + 34 k \begin{array}{l}
(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = (\vec{a} \times \vec{a}) + (\vec{a} \times (-\vec{b})) + (\vec{b} \times \vec{a}) + (\vec{b} \times (-\vec{b})) \\
= (\vec{a} \times \vec{a}) - (\vec{a} \times \vec{b}) + (\vec{b} \times \vec{a}) - (\vec{b} \times \vec{b}) = \vec{0} - (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{b}) - \vec{0} = -2(\vec{a} \times \vec{b}) \\
= -2 \begin{vmatrix} i & j & k \\ 5 & 2 & -1 \\ 1 & -3 & 1 \end{vmatrix} = -2 \begin{pmatrix} 2 & -1 \\ -3 & 1 \end{pmatrix} i - \begin{vmatrix} 5 & -1 \\ 1 & 1 \end{vmatrix} j + \begin{vmatrix} 5 & 2 \\ 1 & -3 \end{vmatrix} k \\
= -2 \left( (2 \cdot 1 - (-3) \cdot (-1))i - (5 \cdot 1 - 1 \cdot (-1))j + (5 \cdot (-3) - 1 \cdot 2)k \right) \\
= -2(-i - 6j - 17k) = 2i + 12j + 34k
\end{array} ( a + b ) × ( a − b ) = ( a × a ) + ( a × ( − b )) + ( b × a ) + ( b × ( − b )) = ( a × a ) − ( a × b ) + ( b × a ) − ( b × b ) = 0 − ( a × b ) − ( a × b ) − 0 = − 2 ( a × b ) = − 2 ∣ ∣ i 5 1 j 2 − 3 k − 1 1 ∣ ∣ = − 2 ( 2 − 3 − 1 1 ) i − ∣ ∣ 5 1 − 1 1 ∣ ∣ j + ∣ ∣ 5 1 2 − 3 ∣ ∣ k = − 2 ( ( 2 ⋅ 1 − ( − 3 ) ⋅ ( − 1 )) i − ( 5 ⋅ 1 − 1 ⋅ ( − 1 )) j + ( 5 ⋅ ( − 3 ) − 1 ⋅ 2 ) k ) = − 2 ( − i − 6 j − 17 k ) = 2 i + 12 j + 34 k
Answer: 2 i + 12 j + 34 k 2i + 12j + 34k 2 i + 12 j + 34 k
www.AssignmentExpert.com
#340153 on Dec 2023