If = 64 then is

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If = 64 then is

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TNUSRB SI 2010 Official Paper (Held on 04 July 2010)

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32
8
128
0

Answer 1

Explanation:

The scalar triple product of three vectors a⃗ a→ $\vec a$ , b⃗ b→ $\vec b$ , and c⃗ c→ $\vec c$ is denoted as [a⃗ ,b⃗ ,c⃗ ][a→,b→,c→] $[\vec a, \vec b, \vec c]$ and defined by:

[a⃗ ,b⃗ ,c⃗ ]=a⃗ ⋅(b⃗ ×c⃗ )=b⃗ ⋅(c⃗ ×a⃗ )=c⃗ ⋅(a⃗ ×b⃗ )[a→,b→,c→]=a→⋅(b→×c→)=b→⋅(c→×a→)=c→⋅(a→×b→)

$ [\vec a, \vec b, \vec c] = \vec a \cdot (\vec b \times \vec c) = \vec b \cdot (\vec c \times \vec a) = \vec c \cdot (\vec a \times \vec b)$

This product is equal to the determinant of a 3x3 matrix whose rows (or columns) are the components of the vectors a⃗ a→ $\vec a$ , b⃗ b→ $\vec b$ , and c⃗ c→ $\vec c$ .

Given that [a⃗ ×b⃗ ,b⃗ ×c⃗ ,c⃗ ×a⃗ ]=64[a→×b→,b→×c→,c→×a→]=64 $[\vec a \times \vec b, \vec b \times \vec c, \vec c \times \vec a] = 64$ , we can use the property of cyclic permutations to rewrite this as [a⃗ ,b⃗ ,c⃗ ]2=64[a→,b→,c→]2=64 $[\vec a, \vec b, \vec c]^2 = 64$ . Therefore, [a⃗ ,b⃗ ,c⃗ ]=±8[a→,b→,c→]=±8 $[\vec a, \vec b, \vec c] = \pm 8$ .

However, the options provided do not include −8−8 $-8$ , so we can conclude that [a⃗ ,b⃗ ,c⃗ ]=8[a→,b→,c→]=8 $[\vec a, \vec b, \vec c] = 8$ , which corresponds to option 2.

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If = 64 then is

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