Let the line l: x=(1-y / -2)=(z-3 / lambda), lambda in mathbbR meet the plane P: x+2 y+3 z=4 at the point (alpha, beta, gamma). If the angle between the line l and the plane P is cos ^-1(sqrt((5 / 14))), then alpha+2 beta+6 gamma is equal to ___________.

L: (x-0 / 1)=(y-1 / 2)=(z-3 / lambda) P: x+2 y+3 z=4 Vector parallel to line : langle 1,2, lambdarangle=barb Normal vector to plane P: * sqrt(1^2+2^2+lambda^2) * sqrt(1^2+2^2+3^2) => frac3sqrt(14)=frac1+4+3 lambdasqrt(lambda^2+5) sqrt(14) => lambda=(2 / 3) L_1 equiv (x-0 / 3)=(y-1 / 6)=(z-3 / 2)=mu Any point on line : (3 mu, 6 mu+1,2 mu+3) It lies on P Therefore 3 mu+12 mu+2+6 mu+9=4 \ => mu=(-1 / 3) Hence, alpha=3 mu=-1, beta=6 mu+1=-1, gamma=2 mu+3=(7 / 3) Now, alpha+2 beta+6 gamma=11

Let the line l xfrac 1y2frac z3lambda lambda R meet the... | null (Mathematics)

L: \frac{x-0}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda}

P: x+2 y+3 z=4

Vector parallel to line :

\langle 1,2, \lambda\rangle=\bar{b}

Normal vector to plane

P:<1,2,3\rangle=\bar{n}

Angle between plane and line is

\theta

Then,

\sin \theta=\frac{<1,2, \lambda>\cdot<1,2,3>}{\sqrt{1^2+2^2+\lambda^2} \cdot \sqrt{1^2+2^2+3^2}}

\Rightarrow \frac{3}{\sqrt{14}}=\frac{1+4+3 \lambda}{\sqrt{\lambda^2+5} \sqrt{14}} \Rightarrow \lambda=\frac{2}{3}

L_1 \equiv \frac{x-0}{3}=\frac{y-1}{6}=\frac{z-3}{2}=\mu

Any point on line :

(3 \mu, 6 \mu+1,2 \mu+3)

It lies on P

\begin{aligned} & \therefore 3 \mu+12 \mu+2+6 \mu+9=4 \\\\ & \Rightarrow \mu=\frac{-1}{3} \end{aligned}

Hence,

\alpha=3 \mu=-1, \beta=6 \mu+1=-1, \gamma=2 \mu+3=\frac{7}{3}

Now,

\alpha+2 \beta+6 \gamma=11

Explanation

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