Std 11, Vectors, 07 cross product, continued

Let us take one more example on cross product. Write down the problem, find the value of A cross B if A and B vectors are together, if we already know A and B vector then lets us find the value of A cross B. okay. Let’s try. We know that A cross B, we can find the value by using the technique of determinant. In determinant we write I cap minus J cap plus K cap. A is the first vector, so let us write the coefficient of the first vector one two and three , B is the second vector six minus two and minus three, lets us take a break , is there a doubt in writing this determinant? Anyone? Sure, if you don’t get it in your exams then it won’t be right.

So now let us find the coefficient of I cap, when we are calculating the value of I cap, this row and this column will be removed, and probably I must have definitely told you; you have to take care of this cross; two into minus three is minus six; minus two into minus three is minus six, isn’t that right? This plus minus errors you have to see sincerely, second minus J cap this row and this column will disappear one minus three product will be minus three, minus three into six is equal to eighteen plus K cap, this row and this column will not be used only these four values are used one into minus two, minus six into two twelve when you solve it, it will become zero I cap plus twenty one J cap minus fourteen K cap so this will be the result of our A cross B.

Let us take one more problem on the cross product.

Let us consider two vectors A and B, if A dot B is equal to zero then what will be the value of magnitude of A cross B, if magnitude of A vector is two and magnitude of B vector is three? So what do we have here is A dot B is equal to zero, so we need to find the value of magnitude of A cross B where in you know the value of A and B. Okay. Done.

Let us try this, if A cross B is equal to zero, what we can find from this particular equation, the angle between the vectors is ninety degrees or pi by two, read out, the magnitude of A cross B is AB sine theta, you know that formula, magnitude of A cross B or the magnitude of their result, and you are creating the cross product. So this will become magnitude of first vector, magnitude of second vector and the angle between the two which is pi by two, and that is equal to six. Any doubts?

Try this particular problem if A dot B is equal to the magnitude of A cross B then find the angle between the A vector and B vector.

A dot B is equal to the magnitude of A cross B let’s try.