Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c.

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Solution

The following figure shows a parallelepiped with sides a, b, c and origin at O.

The expression for volume V of a parallelepiped is given as,

V=abc(1)

Let n ^ be the unit vector perpendicular to vectors b and c, therefore n ^ and a will have the same direction.

The cross product of vector b and c is given as,

b×c=( bcsinθ ) n ^

Here, θ is the angle between b and c.

Substitute the values in the above expression.

b×c=cbsin( 90 ∘ ) n ^ =bc n ^

The dot product of a with b×c is given as,

a⋅( b×c )

Substitute the values in the above expression.

a⋅( b×c )=a⋅( ( bc ) n ^ ) =abccos θ ′

Here, the value of θ ′ will be 0°as the direction of n ^ and a is same.

Substitute the values in the above expression.

a⋅( b×c )=abccos0° =abc [ Since cos0°=1 ] (2)

Hence, it is proved that volume formed a, b and c is equal to the magnitude of a⋅( b×c ).

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Show that a. (b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors, a, b and c.

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a . (b \times c)

is equal in magnitude to the volume of the parallelopiped formed on the three vectors,

a, b

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and

| \to a | = x, | \to b | = y, | \to c | = z

Then the volume of the parallelepiped whose three coterminous edges are

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is_

Q. If

V

is the volume of parallelepiped formed by the vectors

\to a, \to b, \to c

as three coterminous edges is

27

cubic units, then the volume of the parallelepiped having

\to α = \to a + \to 2 b - \to c, \to β = \to a - \to b

and

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as three coterminous edges is

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