Physics, asked by siddharth0017, 9 months ago

Prove that associative property is not valid for  vector product by using following vectors

P = 7i - 6j + 2k and Q = 2i  - j + 3k

Answers

Answered by Anonymous
8

&lt;!DOCTYPE html&gt;</p><p>&lt;html&gt;</p><p>&lt;head&gt;</p><p>&lt;style&gt;</p><p>#grad1 {</p><p>height: 200px;</p><p>background-color: red; /* For browsers that do not support gradients */</p><p>background-image: linear-gradient(to right, red, orange, yellow, green, blue, indigo, violet); /* Standard syntax (must be last) */</p><p>}</p><p>&lt;/style&gt;</p><p>&lt;/head&gt;</p><p>&lt;body&gt;</p><p></p><p>&lt;div id="grad1" style="text-align:center;margin:auto;color:White;font-size:30px;font-weight:bold"&gt;</p><p>&lt;br&gt; </p><p>&lt;/br&gt; ☺️ HEY MATE ☺️</p><p>&lt;/div&gt;</p><p></p><p>&lt;/body&gt;</p><p>&lt;/html&gt;

It should not be confused with the dot product (projection product). together with the cross product is an algebra over the real numbers, which is neither commutative nor associative, but is a Lie algebra with the cross product being the Lie bracket. vectors to produce a vector perpendicular to all of them.

Answered by hannjr
0

Answer:

One way to evaluate the vector cross product is by using a determinant:

                 i       j     k

P X Q =    7     -6     2

                2     -1      3

                i       j       k

Q X P =   2      -1      3

                7      -6    2

You can evaluate these determinants or use the rule that the interchange of two rows in a determinant causes a change in the sign of the determinant.

Similar questions