How to Find Vector Product or Cross Product of Two Vectors :
Here we are going to see how to find vector product or cross product of two vectors.
Working rule to find the cross product
Let
Question 1 :
Find the magnitude of a vector x b vector, if a vector = 2i vector + j vector + 3k vector and b vector = 3i vector+ 5j vector - 2k vector
Solution :
= i vector[-2 - 15] - j vector[-4 -9] + k vector[10-3]
= i vector[-17] - j vector[-13] + k vector[7]
a x b = -17 i vector + 13 j vector + 7 k vector
In order to find its magnitude, we have to take square root and find the sum of coefficients of i, j and k.
|a x b| = √(-17)^{2} + 13^{2} + 7^{2}
= √(289 + 169 + 49)
= √507
Question 2 :
Show that
Solution :
a x (b + c) = a x b + a x c ---(1)
b x (c + a) = b x c + b x a ---(2)
c x (a + b) = c x a + c x b ---(3)
(1) + (2) + (3)
a x (b + c) + b x (c + a) + c x (a + b)
= a x b + a x c + b x c + b x a + c x a + c x b
Since commutative property is not applicable in cross product,
= a x b + a x c + b x c - a x b - a x c - b x c
= 0
Hence it is proved.
Question 3 :
Find the vectors of magnitude 10√3 that are perpendicular to the plane which contains i vector + 2j vector + k vector and i vector + 3j vector + 4k vector
Solution :
Let a vector = i vector + 2j vector + k vector
b vector = i vector + 3j vector + 4k vector
required vector perpendicular to given vectors
= ± μ [(a x b)/ |a x b|]
= i[8-3] - j[4-1] + k[3-2]
a x b = 5i - 3j + k
|a x b| = √5^{2} + (-3)^{2} + 1^{2}
= √(25+9+1)
= √35
Required vector = ± (10√3/√35) (5i - 3j + k)
= ± (10√3/√35) (5i - 3j + k)
After having gone through the stuff given above, we hope that the students would have understood,"Properties of Scalar Product or Dot Product"
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