If -k-k-k-0 and a-x-y-zk- then 8x3-xy-zx is equal to

Î×((â ĵ)×î)+ĵ×((â k̂)×ĵ)+k̂×((â î)×k̂)= 0 ⇒ nbsp; {(î·î)(a ĵ) (î· nbsp; (a ĵ))î} nbsp; + {(ĵ·ĵ)(a k̂) (ĵ· (a k̂))ĵ} + {(k̂·k̂)(a î) (k̂· (a î))k̂}=0 ...

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Byju's Answer

Standard XII

Mathematics

Vector Triple Product

If î×â-ĵ×î+ĵ×...

Question

If $^i \times ((^a -^j) \times^i) +^j \times ((^a -^k) \times^j) +^k \times ((^a -^i) \times^k) = \to 0$ and $\to a = x^i + y^j + z^k,$ then $8 (x^{3} - x y + z x)$ is equal to

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Solution

$^i \times ((^a -^j) \times^i) +^j \times ((^a -^k) \times^j) +^k \times ((^a -^i) \times^k) = \to 0$
$⇒{(^i⋅^i)(→a−^j)−(^i⋅(→a−^j))^i} +{(^j⋅^j)(→a−^k)−(^j⋅(→a−^k))^j} +{(^k⋅^k)(→a−^i)−(^k⋅(→a−^i))^k}=→0$
$\Rightarrow \to a -^j - (^i \cdot \to a)^i + \to a -^k - (^j \cdot \to a)^j + \to a -^i - (^k \cdot \to a)^k = \to 0$
$\Rightarrow 3 \to a - (^i +^j +^k) -^a = \to 0$
$\Rightarrow 2 \to a =^i +^j +^k$
$\Rightarrow^a = \frac{1}{2} (^i +^j +^k) = x^i + y^j + z^k$
$∴ x = y = z = \frac{1}{2}$
$8 (x^{3} - x y + z x) = (x^{3} - x^{2} + x^{2}) = 8 \times \frac{1}{8} = 1$

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If ^i×((^a−^j)×^i)+^j×((^a−^k)×^j)+^k×((^a−^i)×^k)=→0 and →a=x^i+y^j+z^k, then 8(x3−xy+zx) is equal to

If $^i \times ((^a -^j) \times^i) +^j \times ((^a -^k) \times^j) +^k \times ((^a -^i) \times^k) = \to 0$ and $\to a = x^i + y^j + z^k,$ then $8 (x^{3} - x y + z x)$ is equal to