Let b -4 -6 k- and c -2 -7 -10 k- If a is a unit vector and the scalar triple product - a b c- has the greatest value- then a is equal toA- 1-3-k-B- 1-5-2-2k-C- 1-32- i -2 -k-D- 1-593- i -7 -k-

The correct option is C 1/3 (2̂i + 2 ĵ k̂) For [a⃗b⃗c⃗] to be greatest, a⃗ must be perpendicular to both b⃗ and c⃗ i.e. collinear with b⃗×c⃗. b⃗×c⃗ = î am ...

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Byju's Answer

Standard XII

Mathematics

AM,GM,HM Inequality

Let b =-î+4 ĵ...

Question

Let $\to b = -^i + 4^j + 6^k a n d \to c = 2^i - 7^j - 10^k$ . If $\to a$ is a unit vector and the scalar triple product $[\to a \to b \to c]$ has the greatest value, then $\to a$ is equal to

$\frac{1}{\sqrt{3}} (^i +^j +^k)$

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$\frac{1}{\sqrt{5}} (\sqrt{2}^i +^j + \sqrt{2}^k)$

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$\frac{1}{3} (^2 i + 2^j -^k)$

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$\frac{1}{\sqrt{59}} (^3 i + 7^j -^k)$

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Solution

The correct option is C
$\frac{1}{3} (^2 i + 2^j -^k)$

For $[\to a \to b \to c]$ to be greatest, $\to a$ must be perpendicular to both $\to b$ and $\to c$ i.e. collinear with $\to b \times \to c .$

$\to b \times \to c = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k - 1 & 4 & 6 2 & - 7 & - 10 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2^i + 2^j -^k \to a = \frac{1}{3} (2^i + 2^j -^k)$

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Similar questions

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Let →b=−^i+4^j+6^k and →c=2^i−7^j−10^k. If →a is a unit vector and the scalar triple product [→a→b→c] has the greatest value, then →a is equal to

The correct option is C 13(^2i+2^j−^k) For [→a→b→c] to be greatest, →a must be perpendicular to both →b and →c i.e. collinear with →b×→c. →b×→c=∣∣ ∣ ∣∣^i^j^k−1462−7−10∣∣ ∣ ∣∣=2^i+2^j−^k→a=13(2^i+2^j−^k)

Let $\to b = -^i + 4^j + 6^k a n d \to c = 2^i - 7^j - 10^k$ . If $\to a$ is a unit vector and the scalar triple product $[\to a \to b \to c]$ has the greatest value, then $\to a$ is equal to