Consider the set of eight vectors V-a-b-ck- a- b- c-1- 1- Three non-coplanar vector can be chosen from V in 2p ways- Then p is

V={ aî +bĵ +ck̂ :a,b,cϵ{ 1,1 } #160; } =^ 8 C _ 3 = 8! 3!5! = 8× 7× 6 6 =8=2^3 This can be done in 2 ^ 3 ways ∴ p=3 ...

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Standard XII

Mathematics

Cross Product of Two Vectors

Consider the ...

Question

Consider the set of eight vectors $V={a^i+b^j+c^k:a,b,c∈{−1,1}}$ . Three non-coplanar vector can be chosen from $V$ in $2^{p}$ ways. Then $p$ is

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Solution

$V={a^i+b^j+c^k:a,b,cϵ{−1,1}}$
$=^{8} C_{3} = \frac{8!}{3! 5!} = \frac{8 \times 7 \times 6}{6} = 8 = 2^{3}$
This can be done in $2^{3}$ ways
$∴ p = 3$

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Consider the set of eight vectors V={a^i+b^j+c^k:a,b,c∈{−1,1}}. Three non-coplanar vector can be chosen from V in 2p ways. Then p is

V={a^i+b^j+c^k:a,b,cϵ{−1,1}}=8C3=8!3!5!=8×7×66=8=23This can be done in 23 ways∴p=3

Consider the set of eight vectors $V={a^i+b^j+c^k:a,b,c∈{−1,1}}$ . Three non-coplanar vector can be chosen from $V$ in $2^{p}$ ways. Then $p$ is

$V={a^i+b^j+c^k:a,b,cϵ{−1,1}}$
$=^{8} C_{3} = \frac{8!}{3! 5!} = \frac{8 \times 7 \times 6}{6} = 8 = 2^{3}$
This can be done in $2^{3}$ ways
$∴ p = 3$