Chapter 9: Q. 61 (page 628)

Prove property ( $5$ ).

Short Answer

Expert verified

We had proved $u \times (v + w) = (u \times v) + (u \times w)$

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

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

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

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

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

Step 3. Value of the Determinant

On Solving the determinant, we get,

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

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

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

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

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

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

$= (u \times v) + (u \times w)$

Hence proved.

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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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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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Pure Maths

Logic and Functions

Decision Maths

Geometry

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Prove property ( $5$ ).