If angle between planes $\vec{r}\cdot(3\hat{i}-6\hat{j}+2\hat{k})=4$ and $\vec{r}\cdot(2\hat{i}-2\hat{j}-\hat{k})=7$ is $\theta$ then $cos\theta$ equals  Option 1) $\frac{16}{21}$ Option 2) $\frac{17}{21}$ Option 3) $\frac{19}{21}$ Option 4) $\frac{13}{21}$

H Himanshu

As we learned

Angle between two planes (vector form) -

Let the two planes be  $\vec{r}\cdot \vec{n}= d\: and\: \vec{r}\cdot \vec{n_{1}}= d_{1}$ then the angle between them is defined as the andgle between their normals

$\cos \Theta = \frac{\vec{n}\cdot \vec{n_{1}}}{\left | \vec{n} \right |\left | \vec{n_{1}} \right |}$

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Here, $\vec{n}=3\hat{i}-6\hat{j}+2\hat{k}$ and $\vec{n_1}=2\hat{i}-2\hat{j}-\hat{k}$

$\therefore cos\theta=\frac{\vec{n}\cdot{n_1}}{|\vec{n}||\vec{n_1}|}=\frac{6+12-2}{\sqrt{49}\sqrt{9}}=\frac{16}{21}$

Option 1)

$\frac{16}{21}$

Option 2)

$\frac{17}{21}$

Option 3)

$\frac{19}{21}$

Option 4)

$\frac{13}{21}$

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