The angle between the planesr→⋅(2i^−j^+k^)=6 and r→⋅(i^+j^+2k^)=5, is

We know that the angle between the planes r→⋅n1→=d1 and r→⋅n→2=d2 is given by cos⁡0=n→1⋅n→2|n1→|n→2Here, n→1=2i^−j^+k^ and n→2=i^+j^+2k^∴ cos⁡θ=(2i^−j^+k^)⋅(i^+j^+2k^)|2i^−j^+k^||i^+j^+2k^|=12⇒ θ=π/3IF a1x+b3y+c1z+d1=0 and a2x+b2y+c2z+d2=0 are Cartesian equations of two planes, then vectors normal to them are n→1=a1i^+b1j^+c1k^ and n→2=a2i^+b2j^+c2k^respectively. Therefore, the angle 0 between the planes is given bycos⁡θ=n→1⋅n→2n→1n→2⇒ cos⁡θ=a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22If the planes are perpendicular, then n1 → and n2→ are perpendicular. ∴ n1→⋅n2→=0⇒a1a2+b1b2+c1c2=0If the planes are parallel, then n1 → and n2→are parallel∴ a1a2=b1b2=c1c2

The angle between the planesr→⋅(2i^−j^+k^)=6 and r→⋅(i^+j^+2k^)=5, is

We know that the angle between the planes r→⋅n1→=d1 and r→⋅n→2=d2 is given by cos⁡0=n→1⋅n→2|n1→|n→2Here, n→1=2i^−j^+k^ and n→2=i^+j^+2k^∴ cos⁡θ=(2i^−j^+k^)⋅(i^+j^+2k^)|2i^−j^+k^||i^+j^+2k^|=12⇒ θ=π/3IF a1x+b3y+c1z+d1=0 and a2x+b2y+c2z+d2=0 are Cartesian equations of two planes, then vectors normal to them are n→1=a1i^+b1j^+c1k^ and n→2=a2i^+b2j^+c2k^respectively. Therefore, the angle 0 between the planes is given bycos⁡θ=n→1⋅n→2n→1n→2⇒ cos⁡θ=a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22If the planes are perpendicular, then n1 → and n2→ are perpendicular. ∴ n1→⋅n2→=0⇒a1a2+b1b2+c1c2=0If the planes are parallel, then n1 → and n2→are parallel∴ a1a2=b1b2=c1c2

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The angle between the planes $\vec{r} \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 6$ and $\vec{r} \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 5$ , is

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$\frac{π}{3}$

$\frac{2 π}{3}$

$\frac{π}{6}$

$\frac{5 π}{6}$

answer is A.

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Detailed Solution

We know that the angle between the planes $\vec{r} \cdot \vec{n_{1}} = d_{1} and \vec{r} \cdot {\vec{n}}_{2} = d_{2}$ is given by

$\cos 0 = \frac{{\vec{n}}_{1} \cdot \vec{n} 2}{| \vec{n 1} | |{\vec{n}}_{2}|}$

Here, ${\vec{n}}_{1} = 2 \hat{i} - \hat{j} + \hat{k} and {\vec{n}}_{2} = \hat{i} + \hat{j} + 2 \hat{k}$

$∴ \cos θ = \frac{(2 \hat{i} - \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} + 2 \hat{k})}{| 2 \hat{i} - \hat{j} + \hat{k} | | \hat{i} + \hat{j} + 2 \hat{k} |} = \frac{1}{2}$

$\Rightarrow θ = π / 3$

IF $a_{1} x + b_{3} y + c_{1} z + d_{1} = 0 and a_{2} x + b_{2} y + c_{2} z + d_{2} = 0$ are Cartesian equations of two planes, then vectors normal to them are

${\vec{n}}_{1} = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k} and {\vec{n}}_{2} = a_{2} \hat{i} + b_{2} \hat{j} + c_{2} \hat{k}$

respectively.

Therefore, the angle 0 between the planes is given by

$\begin{array}{l} \cos θ = \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{|{\vec{n}}_{1}| |{\vec{n}}_{2}|} \\ \Rightarrow \cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}} \end{array}$