Find angle using sin θ = |b⃗·n⃗| / (|b⃗||n⃗|).

Angle Between a Line and a Plane

Question

A line is given in vector form as $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{k})$ . Find the angle it makes with the plane $\vec{r} \cdot (\hat{j} + \hat{k}) = 4$ .

(CBSE 2025 Sample Paper)

Solution — Step by Step

From the line equation, the direction vector is $\vec{b} = \hat{i} + \hat{k}$ , so direction ratios are $(1, 0, 1)$ .

From the plane $\vec{r} \cdot (\hat{j} + \hat{k}) = 4$ , the normal vector is $\vec{n} = \hat{j} + \hat{k}$ , so direction ratios are $(0, 1, 1)$ .

The angle $\theta$ between a line and a plane satisfies:

\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|}

Why $\sin$ and not $\cos$ ? The angle between a line and a plane is measured from the plane, not from the normal. That makes it the complement of the angle between the line and the normal — and $\sin(90° - \phi) = \cos\phi$ , which flips the formula.

\vec{b} \cdot \vec{n} = (1)(0) + (0)(1) + (1)(1) = 1

|\vec{b}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}

|\vec{n}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}

\sin \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}

\boxed{\theta = 30°}

Why This Works

The key insight is that a plane has infinitely many lines lying inside it — and the angle between our line and the plane is measured relative to that surface, not relative to the normal sticking out of it.

If $\phi$ is the angle between the line and the normal, then $\theta = 90° - \phi$ . We know $\cos\phi = \frac{|\vec{b}\cdot\vec{n}|}{|\vec{b}||\vec{n}|}$ , and since $\sin\theta = \cos(90° - \theta) = \cos\phi$ , we get the $\sin$ formula directly.

We always take the absolute value of $\vec{b} \cdot \vec{n}$ because the angle between a line and a plane is defined to lie between $0°$ and $90°$ — reversing the line’s direction shouldn’t change the answer.

Alternative Method

We can verify using the complement directly.

The angle $\phi$ between the line $\vec{b} = (1, 0, 1)$ and the normal $\vec{n} = (0, 1, 1)$ :

\cos\phi = \frac{|\vec{b}\cdot\vec{n}|}{|\vec{b}||\vec{n}|} = \frac{1}{\sqrt{2}\cdot\sqrt{2}} = \frac{1}{2} \implies \phi = 60°

Then the angle with the plane is:

\theta = 90° - \phi = 90° - 60° = 30°

Same answer, slightly longer. In a 3-mark CBSE question or JEE Main, use the $\sin\theta$ formula directly — saves 30 seconds and avoids the two-step process.

Common Mistake

The most common error here is writing $\cos\theta = \frac{|\vec{b}\cdot\vec{n}|}{|\vec{b}||\vec{n}|}$ — that’s the formula for the angle between two lines. For a line and a plane, the formula uses $\sin\theta$ . This single substitution error drops your answer from $30°$ to $60°$ , and in CBSE marking, the method marks go too since the setup is wrong from the start.

Quick memory hook for exams: Line-Line → cos, Line-Plane → sin. When a plane enters the picture, the formula flips. If you remember this one distinction, you will never mix up the two formulas again — and this topic has solid weightage in both CBSE Class 12 and JEE Main.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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