Chapter 9: Q. 60 (page 628)

Prove property $(3)$

Short Answer

Expert verified

We had proved $u \times v = - (v \times u)$

Step by step solution

Step 1. Assumption Of Vectors

Let us assume that the two vectors $u, v$ be:

$u = a_{1} i + b_{1} j + c_{1} k and v = a_{2} i + b_{2} j + c_{2} k$

Step 2. Solve Left Hand Side

Now, we will solve the lest hand side of the equation,

$u \times v = |\begin{array}{l} i & j & k \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}|$

$= |\begin{array}{l} b_{1} & c_{1} \\ b_{2} & c_{2} \end{array}| i - |\begin{array}{l} a_{1} & c_{1} \\ a_{2} & c_{2} \end{array}| j + |\begin{array}{l} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}| k$

$= (b_{1} c_{2} - b_{2} c_{1}) i - (a_{1} c_{2} - a_{2} c_{1}) j + (a_{1} b_{2} - a_{2} b_{1}) k$

$= - (b_{2} c_{1} - b_{1} c_{2}) i + (a_{2} c_{1} - a_{1} c_{2}) j - (a_{2} b_{1} - a_{1} b_{2}) k$

$= - [(b_{2} c_{1} - b_{1} c_{2}) i - (a_{2} c_{1} - a_{1} c_{2}) j + (a_{2} b_{1} - a_{1} b_{2}) k]$

Step 3. Find the Determinant Form

The above equation can be written in determinant form as,

$= - [\begin{array}{l} b_{2} & c_{2} \\ b_{1} & c_{1} \end{array} | i - | \begin{array}{l} a_{2} & c_{2} \\ a_{1} & c_{1} \end{array} | j + | \begin{matrix} a_{2} & b_{2} \\ a_{1} & b_{1} \end{matrix} ∣ k]$

$= - |\begin{matrix} i & j & k \\ a_{2} & b_{2} & c_{2} \\ a_{1} & b_{1} & c_{1} \end{matrix}|$

$= - (v \times u)$

Hence Proved.

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Prove property $(3)$

Short Answer

Step by step solution

Step 1. Assumption Of Vectors

Step 2. Solve Left Hand Side

Step 3. Find the Determinant Form

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Prove property(3)

Short Answer

Step by step solution

Step 1. Assumption Of Vectors

Step 2. Solve Left Hand Side

Step 3. Find the Determinant Form

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Discrete Mathematics

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Theoretical and Mathematical Physics

Probability and Statistics

Pure Maths

Study anywhere. Anytime. Across all devices.

Prove property $(3)$