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Mathematics · Ch 6 — Applications of Vector Algebra

Angle Between a Line and a Plane

6.8.11

Angle Between a Line and a Plane

The angle between a line and a plane is the complement of the angle between the line's direction and the plane's normal — because a line lying flat IN the plane makes a 0°0° angle with the plane while its direction is perpendicular (90°90°) to the normal, and vice versa for a line along the normal.

Let r⃗=a⃗+tb⃗\vec r=\vec a+t\vec b be the line and r⃗⋅n⃗=p\vec r\cdot\vec n=p the plane. If θ\theta is the acute angle between the LINE and the PLANE, then the acute angle between n⃗\vec n and b⃗\vec b is π2−θ\dfrac\pi2-\theta, so

cos⁡(π2−θ)=b⃗⋅n⃗∣b⃗∣∣n⃗∣ ⟹ sin⁡θ=b⃗⋅n⃗∣b⃗∣∣n⃗∣,\cos\left(\frac\pi2-\theta\right)=\frac{\vec b\cdot\vec n}{|\vec b||\vec n|}\ \Longrightarrow\ \sin\theta=\frac{\vec b\cdot\vec n}{|\vec b||\vec n|},

giving

θ=sin⁡−1(∣b⃗⋅n⃗∣b⃗∣∣n⃗∣∣).\boxed{\theta=\sin^{-1}\left(\left|\frac{\vec b\cdot\vec n}{|\vec b||\vec n|}\right|\right).}

Cartesian form. For the line x−x1a1=y−y1b1=z−z1c1\dfrac{x-x_1}{a_1}=\dfrac{y-y_1}{b_1}=\dfrac{z-z_1}{c_1} and the plane ax+by+cz=pax+by+cz=p:

θ=sin⁡−1(∣aa1+bb1+cc1∣a12+b12+c12 a2+b2+c2).\theta=\sin^{-1}\left(\frac{|aa_1+bb_1+cc_1|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a^2+b^2+c^2}}\right).

Remark. …

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