If \(A_{1}, A_{2}\), and \(A_{3}\) are three arbitrary vectors that are not coplanar, show that any arbitrary dyadic T can always be expressed as the sum of three dyads $$ T=A_{1} B_{1}+A_{2} B_{2}+A_{2} B_{2} $$ by suitably choosing the three vectors \(\mathbf{B}_{1}, \mathbf{B}_{2}\), and \(\mathbf{B}_{2}\).

Short Answer

Expert verified
Question: Prove that any arbitrary dyadic T can always be expressed as the sum of three dyads by choosing suitable three vectors B1, B2, and B3, if given three arbitrary non-coplanar vectors A1, A2, and A3. Answer: By expressing B1, B2, and B3 as linear combinations of A1, A2, and A3, we can write the sum of three dyads. Through the distribution of A1, A2, and A3 over the expressions, we can create an expression for T as a sum of dyads formed by the basis vectors A1, A2, and A3. By matching the dyads from this expression with the given expression for T and selecting appropriate scalar coefficients αij, we can obtain any dyadic T. Thus, we have proved that any arbitrary dyadic T can always be expressed as the sum of three dyads by suitably selecting the three vectors B1, B2, and B3.

Step by step solution

01

Write down the expression to find T

To begin with, we will write down the expression that we are supposed to prove: $$ T=A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3} $$
02

Rewrite expression using Linear combinations

Since A1, A2, and A3 are arbitrary non-coplanar vectors, they span a three-dimensional space over the field of scalars. Therefore, we can express any vector in terms of these three vectors (also known as the basis vectors). Let's express B1, B2, and B3 as linear combinations of A1, A2, and A3 as follows: $$ \mathbf{B}_{1} = \alpha_{11} \mathbf{A}_{1} + \alpha_{21} \mathbf{A}_{2} + \alpha_{31} \mathbf{A}_{3}\\ \mathbf{B}_{2} = \alpha_{12} \mathbf{A}_{1} + \alpha_{22} \mathbf{A}_{2} + \alpha_{32} \mathbf{A}_{3}\\ \mathbf{B}_{3} = \alpha_{13} \mathbf{A}_{1} + \alpha_{23} \mathbf{A}_{2} + \alpha_{33} \mathbf{A}_{3} $$ where αij are scalar coefficients.
03

Substitute B1, B2, and B3 in the expression for T

Now substitute the expressions for B1, B2, and B3 from step 2 into the expression for T: $$ T= A_{1}(\alpha_{11} \mathbf{A}_{1} + \alpha_{21} \mathbf{A}_{2} + \alpha_{31} \mathbf{A}_{3})+A_{2}(\alpha_{12} \mathbf{A}_{1} + \alpha_{22} \mathbf{A}_{2} + \alpha_{32} \mathbf{A}_{3})+A_{3}(\alpha_{13} \mathbf{A}_{1} + \alpha_{23} \mathbf{A}_{2} + \alpha_{33} \mathbf{A}_{3}) $$
04

Distribute A1, A2, and A3 over the expressions

Now distribute A1, A2, and A3 over the expressions inside the parentheses: $$ T= A_{1}(\alpha_{11} \mathbf{A}_{1}) + A_{1}(\alpha_{21} \mathbf{A}_{2}) + A_{1}(\alpha_{31} \mathbf{A}_{3})+A_{2}(\alpha_{12} \mathbf{A}_{1}) + A_{2}(\alpha_{22} \mathbf{A}_{2}) + A_{2}(\alpha_{32} \mathbf{A}_{3})+A_{3}(\alpha_{13} \mathbf{A}_{1}) + A_{3}(\alpha_{23} \mathbf{A}_{2}) + A_{3}(\alpha_{33} \mathbf{A}_{3}) $$ At this stage, we have an expression for T as a sum of dyads formed by the basis vectors A1, A2, and A3.
05

Match dyads against the given expression

We can match the dyads from the above expression with the given expression for T: $$ T=A_{1}B_{1}+A_{2}B_{2}+A_{3}B_{3} $$ By doing so, we can see that by selecting the appropriate scalar coefficients αij, we can obtain any dyadic T. Hence, we have shown that an arbitrary dyadic T can always be expressed as the sum of three dyads by suitably choosing the three vectors B1, B2, and B3.

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