If |a¯|=4,|b¯|=5,|a¯−b¯|=3 and θ is the angle between the vectors a¯ and b¯, then tan2θ=

Now, |a−b|2=|a|2+|b|2−2a.b⇒(3)2=42+52−2a.b⇒9=16+25−2a.b⇒2a.b=32 ⇒a.b=16Now, we know thatcosθ=a.b|a|.|b|=164.5cosθ=45tanθ=34⇒tan2θ=916

If |a¯|=4,|b¯|=5,|a¯−b¯|=3 and θ is the angle between the vectors a¯ and b¯, then tan2θ=

Now, |a−b|2=|a|2+|b|2−2a.b⇒(3)2=42+52−2a.b⇒9=16+25−2a.b⇒2a.b=32 ⇒a.b=16Now, we know thatcosθ=a.b|a|.|b|=164.5cosθ=45tanθ=34⇒tan2θ=916

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If $| \bar{a} | = 4, | \bar{b} | = 5, | \bar{a} - \bar{b} | = 3$ and $θ$ is the angle between the vectors $\bar{a}$ and $\bar{b}$ , then $\tan^{2} θ =$

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$\frac{3}{4}$

$\frac{9}{16}$

$\frac{16}{9}$

$\frac{4}{3}$

answer is D.

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Detailed Solution

Now, $| a - b |^{2} = | a |^{2} + | b |^{2} - 2 a . b$
$\begin{array}{l} \Rightarrow {(3)}^{2} = 4^{2} + 5^{2} - 2 a . b \Rightarrow 9 = 16 + 25 - 2 a . b \\ \Rightarrow 2 a . b = 32 \Rightarrow a . b = 16 \end{array}$
Now, we know that
$\begin{array}{l} \cos θ = \frac{a . b}{| a | . | b |} = \frac{16}{4.5} \cos θ = \frac{4}{5} \\ \tan θ = \frac{3}{4} \end{array}$
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$\Rightarrow \tan^{2} θ = \frac{9}{16}$