Vectors a - b and c are of the same length and when taken pairwise- they form equal angles- If a - and b -k- then vector c can beA- -k-B- 1-3-4-3-1-3k- -1-3-4-3-1-3k-D- -2 -k-

The correct options are A î+k̂ nbsp; C 13î+43ĵ 13k̂ nbsp; nbsp;Let c =x î+yĵ+zk̂ Then |a|=|b|=|c| ⇒ x^2+y^2+z^2=2 It is given that the angles between the ...

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Standard XII

Mathematics

Dot Product of Two Vectors

Vectors a , b...

Question

Vectors $\to a, \to b$ and $\to c$ are of the same length and when taken pairwise, they form equal angles. If $\to a =^i +^j$ and $\to b =^j +^k$ , then vector $\to c$ can be

^i +^k

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\frac{1}{3}^i - \frac{4}{3}^j - \frac{1}{3}^k

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- \frac{1}{3}^i + \frac{4}{3}^j - \frac{1}{3}^k

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^i + 2^j +^k

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Solution

The correct options are
A $^i +^k$
C $- \frac{1}{3}^i + \frac{4}{3}^j - \frac{1}{3}^k$
Let $\to c = x^i + y^j + z^k$
Then $| \to a | = | \to b | = | \to c | \Rightarrow x^{2} + y^{2} + z^{2} = 2$
It is given that the angles between the vectors taken in pairs are equal, say $θ$ .
Therefore, $cos θ = \frac{\to a . \to b}{| \to a | | \to b |} = \frac{0 + 1 + 0}{\sqrt{2} \sqrt{2}} = \frac{1}{2}$
$\Rightarrow \frac{\to a . \to c}{| \to a | | \to c |} = \frac{1}{2}$ and $\frac{\to b . \to c}{| \to b | | \to c |} = \frac{1}{2}$
$\Rightarrow \frac{x + y}{2} = \frac{1}{2}$ and $\frac{y + z}{2} = \frac{1}{2}$
$\Rightarrow x + y = 1$ and $y + z = 1$
$\Rightarrow y = 1 - x$ and $z = 1 - y = 1 - (1 - x) = x$

$x^{2} + y^{2} + z^{2} = 2 \Rightarrow x^{2} + (1 - x)^{2} + x^{2} = 2$
$\Rightarrow (3 x + 1) (x - 1) = 0$
$\Rightarrow x = 1, - \frac{1}{3}$
Now, $y = 1 - x$
$\Rightarrow y = 0$ for $x = 1$ and $y = \frac{4}{3}$ for $x = - \frac{1}{3}$
Hence, $\to c =^i +^k$ or $\to c = - \frac{1}{3}^i + \frac{4}{3}^j - \frac{1}{3}^k$

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Dot Product of Two Vectors

Standard XII Mathematics

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Vectors $\to a, \to b$ and $\to c$ are of the same length and when taken pairwise, they form equal angles. If $\to a =^i +^j$ and $\to b =^j +^k$ , then vector $\to c$ can be