Q. 61 Prove property (5).聽... [FREE SOLUTION]

91影视

Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 9: Q. 61 (page 628)

Prove property ( $5$ ).

Short Answer

Expert verified

We had proved $u 脳 (v + w) = (u 脳 v) + (u 脳 w)$

Step by step solution

Step 1. Assumption Of Vectors

Let the three vectors be:

$u = a_{1} i + b_{1} j + c_{1} k, v = a_{2} i + b_{2} j + c_{2} k and w = a_{3} i + b_{3} j + c_{3} k$

Step 2. Calculating One Side Of the equation

The left hand side of the equation is

$u 脳 (v + w) = u 脳 ((a_{2} i + b_{2} j + c_{2} k) + (a_{3} i + b_{3} j + c_{3} k))$

$= u 脳 [(a_{2} + a_{3}) i + (b_{2} + b_{3}) j + (c_{2} + c_{3}) k]$

$= (a_{1} i + b_{1} j + c_{1} k) 脳 [(a_{2} + a_{3}) i + (b_{2} + b_{3}) j + (c_{2} + c_{3}) k]$

$= |\begin{matrix} i & j & k \\ a_{1} & b_{1} & c_{1} \\ a_{2} + a_{3} & b_{2} + b_{3} & c_{2} + c_{3} \end{matrix}|$

Step 3. Value of the Determinant

On Solving the determinant, we get,

$= |\begin{matrix} b_{1} & c_{1} \\ b_{2} + b_{3} & c_{2} + c_{3} \end{matrix}| i - |\begin{matrix} a_{1} & c_{1} \\ a_{2} + a_{3} & c_{2} + c_{3} \end{matrix}| j + |\begin{matrix} a_{1} & b_{1} \\ a_{2} + a_{3} & b_{2} + b_{3} \end{matrix}| k$

$= (b_{1} c_{2} + b_{1} c_{3} - b_{2} c_{1} - b_{3} c_{1}) i - (a_{1} c_{2} + a_{1} c_{3} - a_{2} c_{1} - a_{3} c_{1}) j + (a_{1} b_{2} + a_{1} b_{3} - a_{2} b_{1} - a_{3} b_{1}) k$

$= (b_{1} c_{2} - b_{2} c_{1}) i + (b_{1} c_{3} - b_{3} c_{1}) i - (a_{1} c_{2} - a_{2} c_{1}) j - (a_{1} c_{3} - a_{3} c_{1}) j + (a_{1} b_{2} - a_{2} b_{1}) k + (a_{1} b_{3} - a_{3} b_{1}) k$

$= (b_{1} c_{2} - b_{2} c_{1}) i - (a_{1} c_{2} - a_{2} c_{1}) j + (a_{1} b_{2} - a_{2} b_{1}) k + (b_{1} c_{3} - b_{3} c_{1}) i - (a_{1} c_{3} - a_{3} c_{1}) j + (a_{1} b_{3} - a_{3} b_{1}) k$

$= (|\begin{array}{l} b_{1} & c_{1} \\ b_{2} & c_{2} \end{array}| i - |\begin{array}{l} a_{1} & c_{1} \\ a_{2} & c_{2} \end{array}| j + |\begin{array}{l} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}| k) + (|\begin{array}{l} b_{1} & c_{1} \\ b_{3} & c_{3} \end{array}| i - |\begin{array}{l} a_{1} & c_{1} \\ a_{3} & c_{3} \end{array}| j + |\begin{array}{l} a_{1} & b_{1} \\ a_{3} & b_{3} \end{array}| k)$

$= |\begin{array}{l} i & j & k \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}| + |\begin{array}{l} i & j & k \\ a_{1} & b_{1} & c_{1} \\ a_{3} & b_{3} & c_{3} \end{array}|$

$= (u 脳 v) + (u 脳 w)$

Hence proved.

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91影视

Prove property ( $5$ ).

Short Answer

Step by step solution

Step 1. Assumption Of Vectors

Step 2. Calculating One Side Of the equation

Step 3. Value of the Determinant

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