The Problems (From May 21st post):
1) For a vector A = 3i – 2j – k find the magnitude of the vector product: A X A.
2) A cube has edges of length 2a, with center at 0 and its sides are parallel to the coordinate planes of an xyz coordinate system. What are the position vectors of the corners?
3) Find A X B if A = 2i - 2j -k, and B = i + j + k.
4) Find a vector perpendicular to both of the vectors A = i + j + k and B = i + j
5) Let A = i + j, B = 2i - 3j -k and: C = 4j - 3k
Show that: (A X B) X C = A X (B X C) (Using the determinant method)
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Solutions:
1) The magnitude of the cross product is given by the product of the magnitudes of the two vectors times the sine of the angle between them.
A X B = n ‖A‖ ‖B‖ sin Θ
But if the two vectors are equal, the angle between them is 0, so the sine of the angle is 0, so the magnitude of the cross product is 0, making it the zero vector. Hence:
A X A = ‖A‖ ‖A‖ sin 0 = 0
2) The cube is sketched below:
From the sketch and the dimensions shown we can deduce the position vectors for the corners: + a x i + a y j + a z k
But: a x = a y = a z = a (in length)
So; Position vectors are: + a i + a j + a k
= a (+ i + j + k )
3) A = 2i - 2j -k = a x i + a y j + a z k
Where: a x = 2, a y = -2, a z = -1
And: B = i + j + k = b x i + b y j + b z k
Where: b x = b y = b z = 1
So that:
A X B = (a y b z - a z b y )i + (a z b x - a x b z )j +
(a x b y - a y b x )k
= {(-2) (1) - (-1) (1) } i + {(-1) (1) - (2)(1)}j + {(2)(1) -(-2)(1)}k
= (-2 + 1)i + (-1 - 2)j + (2 +2) k = - i -3j + 4k
4) We need the vector cross product ( A X B ) for A = i + j + k and B = i + j
A = i + j + k = a x i + a y j + a z k
For which: a x = a y = a z = 1
B = i + j = b x i + b y j
For which: b x = b y = 1 and: b z = 0 (i.e. no component)
Then: A X B = ( - a z b y )i + (a z b x ) j +
(a x b y - a y b x )k
= { (-1) (1)} i + {(1) (1)}j + {(1)(1) - (1)(1)} k = - i + j
5) Using the method of determinants:
With: A = i + j, B = 2i - 3j -k and: C = 4j - 3k
For which: a x =1, a y = 1, a z = 0
b x =2, b y = -3, b z = -1
c y = 4, c z = -3
3) Find A X B if A = 2i - 2j -k, and B = i + j + k.
4) Find a vector perpendicular to both of the vectors A = i + j + k and B = i + j
5) Let A = i + j, B = 2i - 3j -k and: C = 4j - 3k
Show that: (A X B) X C = A X (B X C) (Using the determinant method)
--------------------------------------
Solutions:
1) The magnitude of the cross product is given by the product of the magnitudes of the two vectors times the sine of the angle between them.
A X B = n ‖A‖ ‖B‖ sin Θ
But if the two vectors are equal, the angle between them is 0, so the sine of the angle is 0, so the magnitude of the cross product is 0, making it the zero vector. Hence:
A X A = ‖A‖ ‖A‖ sin 0 = 0
2) The cube is sketched below:
From the sketch and the dimensions shown we can deduce the position vectors for the corners: + a x i + a y j + a z k
But: a x = a y = a z = a (in length)
So; Position vectors are: + a i + a j + a k
= a (+ i + j + k )
3) A = 2i - 2j -k = a x i + a y j + a z k
Where: a x = 2, a y = -2, a z = -1
And: B = i + j + k = b x i + b y j + b z k
Where: b x = b y = b z = 1
So that:
A X B = (a y b z - a z b y )i + (a z b x - a x b z )j +
(a x b y - a y b x )k
= {(-2) (1) - (-1) (1) } i + {(-1) (1) - (2)(1)}j + {(2)(1) -(-2)(1)}k
= (-2 + 1)i + (-1 - 2)j + (2 +2) k = - i -3j + 4k
4) We need the vector cross product ( A X B ) for A = i + j + k and B = i + j
A = i + j + k = a x i + a y j + a z k
For which: a x = a y = a z = 1
B = i + j = b x i + b y j
For which: b x = b y = 1 and: b z = 0 (i.e. no component)
Then: A X B = ( - a z b y )i + (a z b x ) j +
(a x b y - a y b x )k
= { (-1) (1)} i + {(1) (1)}j + {(1)(1) - (1)(1)} k = - i + j
5) Using the method of determinants:
With: A = i + j, B = 2i - 3j -k and: C = 4j - 3k
For which: a x =1, a y = 1, a z = 0
b x =2, b y = -3, b z = -1
A X B = [ i j k ]
[ a x a y a z]
[ b x b y b z]
o:
A X B = [ i j k ]
[ 1 1 0]
[ 2 -3 -1]
= i - j - 5 k
Then:
(A X B) X C = [ i j k ]
[ 1 -1 -5 ]
[ 0 4 -3]
Then: (A X B) X C = 23i + 3j + 4k
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