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 + 132 + 72
= √(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| = √52 + (-3)2 + 12
= √(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"
Apart from the stuff given in "Properties of Scalar Product or Dot Product", if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Apr 22, 24 01:36 PM
Apr 21, 24 08:29 PM
Apr 21, 24 01:23 AM