Answer to Question #98724 in Analytic Geometry for jahmal usen

Question

if \\(a=a_{1}i+a_{2}j+a_{3}k\\) and \\(b=b_{1}i+b_{2}j+b_{3}k\\), then ab

Answer 1

The geometric definition of the dot product says that the dot product between two vectors a and b is $a \cdot b = \lVert a \rVert \lVert b \rVert cos \theta$ , where $\theta$ is the angle between vectors a and b.

As a first step, we look at the dot product between standard unit vectors, i.e., the vectors i, j, and k of length one and parallel to the coordinate axes.

Since the standard unit vectors are orthogonal, we immediately conclude that the dot product between a pair of distinct standard unit vectors is zero: $i \cdot j = k \cdot j = i \cdot k = 0$  .

The dot product between a unit vector and itself is also simple to compute. In this case, the angle is zero and $cos \theta= 1$ . Given that the vectors are all of length one, the dot products are $i \cdot i = j \cdot j = k \cdot k = 1$ .

The second step is to calculate the dot product between two three-dimensional vectors

$a =(a_1,a_2,a_3)=a_1i+a_2j+a_3k$ , $b =(b_1,b_2,b_3)=b_1i+b_2j+b_3k$ .

To do this, we simply assert that for any three vectors a, b, and c, and any scalar $\lambda$ : $(\lambda a) \cdot b = \lambda (a \cdot b ) = a \cdot (\lambda b)$ , $(a+b) \cdot c = a \cdot c + b \cdot c$ .

Given these properties and the fact that the dot product is commutative, we can expand the dot product $a \cdot b$ in terms of components, $a \cdot b =(a_1i+a_2j+a_3k) \cdot (b_1i+b_2j+b_3k) = a_1b_1i \cdot i + a_2b_2j \cdot j + a_3b_3k \cdot k + (a_1b_2 + a_2b_1)i \cdot j + (a_1b_3 + a_3b_1)i \cdot k + (a_2b_3+a_3b_2)j \cdot k$

Since we know the dot product of unit vectors, we can simplify the dot product formula to $a \cdot b = a_1b_1+a_2b_2+a_3b_3$ .

Answer: $a \cdot b = a_1b_1+a_2b_2+a_3b_3$ .