Eight Sure 3-Mark Questions for Class 12

This lesson works through eight sure three-mark questions for the Class 12 board exam, covering vectors, conditional probability, definite integrals, and 3D coordinate geometry. Each question is solved fully so you can follow the exact method.

Magnitude of a cross product

Given $|\vec a| = 10$, $|\vec b| = 2$, and $\vec a \cdot \vec b = 12$, find $|\vec a \times \vec b|$.

The dot product gives the cosine, so start there:

$\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|} = \frac{12}{10 \times 2} = \frac{3}{5}.$

Then

$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \tfrac{9}{25}} = \sqrt{\tfrac{16}{25}} = \frac{4}{5}.$

Now use $|\vec a \times \vec b| = |\vec a|\,|\vec b|\,\sin\theta$:

$|\vec a \times \vec b| = 10 \times 2 \times \frac{4}{5} = 16.$

Angle between two vectors

Given $|\vec a \times \vec b| = 1$, $|\vec a| = 2$, and $|\vec b| = 1$, find the angle between $\vec a$ and $\vec b$.

Here the cross product magnitude is given, so use the sine relation:

$\sin\theta = \frac{|\vec a \times \vec b|}{|\vec a|\,|\vec b|} = \frac{1}{2 \times 1} = \frac{1}{2}.$

So

$\theta = \frac{\pi}{6} = 30^\circ.$

The choice of which relation to use depends on whether the dot product or the cross product is given.

Values of lambda for an obtuse angle

Find the values of $\lambda$ for which the angle between $\vec a = 2\lambda^2\,\hat i + 4\lambda\,\hat j + \hat k$ and $\vec b = 7\,\hat i - 2\,\hat j + \lambda\,\hat k$ is obtuse.

An obtuse angle (greater than $90^\circ$) means $\cos\theta < 0$. Since

$\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|},$

and the magnitudes are positive, the condition reduces to $\vec a \cdot \vec b < 0$.

Dot product. Multiply matching components and add:

$\vec a \cdot \vec b = (2\lambda^2)(7) + (4\lambda)(-2) + (1)(\lambda) = 14\lambda^2 - 8\lambda + \lambda = 14\lambda^2 - 7\lambda.$

Solve the inequality. Require $14\lambda^2 - 7\lambda < 0$, that is

$7\lambda\,(2\lambda - 1) < 0.$

This product is negative when $\lambda > 0$ and $2\lambda - 1 < 0$, so

$0 < \lambda < \frac{1}{2}, \qquad \lambda \in \left(0, \tfrac{1}{2}\right).$

Proving a conditional probability lies between 0 and 1

Let $A$ and $B$ be events of the same sample space with $P(B) \neq 0$. Prove that $0 \le P(A \mid B) \le 1$.

By definition,

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$

Upper bound. Since $A \cap B \subseteq B$, we have $P(A \cap B) \le P(B)$. Dividing by $P(B) > 0$,

$\frac{P(A \cap B)}{P(B)} \le 1.$

Lower bound. Probabilities are non-negative, so $P(A \cap B) \ge 0$, and with $P(B) > 0$,

$\frac{P(A \cap B)}{P(B)} \ge 0.$

Combining the two,

$0 \le \frac{P(A \cap B)}{P(B)} \le 1 \quad\Longrightarrow\quad 0 \le P(A \mid B) \le 1.$

Definite integral of sine squared by symmetry

Evaluate $\displaystyle\int_0^{\pi/2} \sin^2 x \, dx$.

Let

$I = \int_0^{\pi/2} \sin^2 x \, dx. \qquad (1)$

Use the property $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ with $a = \tfrac{\pi}{2}$. Since $\sin\left(\tfrac{\pi}{2} - x\right) = \cos x$,

$I = \int_0^{\pi/2} \cos^2 x \, dx. \qquad (2)$

Add (1) and (2):

$2I = \int_0^{\pi/2} \left(\sin^2 x + \cos^2 x\right) dx = \int_0^{\pi/2} 1 \, dx = \Big[x\Big]_0^{\pi/2} = \frac{\pi}{2}.$

Therefore

$I = \frac{\pi}{4}.$

Definite integral by substitution

Evaluate $\displaystyle\int_0^{\pi/2} \frac{\sin x}{1 + \cos^2 x} \, dx$.

Put $t = \cos x$, so $-\sin x \, dx = dt$, that is $\sin x \, dx = -\,dt$. Change the limits: when $x = 0$, $t = \cos 0 = 1$; when $x = \tfrac{\pi}{2}$, $t = \cos\tfrac{\pi}{2} = 0$.

$I = \int_1^0 \frac{-\,dt}{1 + t^2} = -\int_1^0 \frac{dt}{1 + t^2} = -\Big[\tan^{-1} t\Big]_1^0.$

Evaluating,

$I = -\left(\tan^{-1} 0 - \tan^{-1} 1\right) = -\left(0 - \frac{\pi}{4}\right) = \frac{\pi}{4}.$

Finding lambda for three collinear points

If $A(-1, 3, 2)$, $B(-4, 2, -2)$, and $C(5, 5, \lambda)$ are collinear, find $\lambda$.

First write the Cartesian equation of line $AB$ through $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$:

$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}.$

Substituting $A$ and $B$,

$\frac{x + 1}{-4 + 1} = \frac{y - 3}{2 - 3} = \frac{z - 2}{-2 - 2} = \frac{x + 1}{-3} = \frac{y - 3}{-1} = \frac{z - 2}{-4}.$

Clearing the common negative signs,

$\frac{x + 1}{3} = \frac{y - 3}{1} = \frac{z - 2}{4}. \qquad (1)$

Since $C$ is collinear, it lies on this line. Put $x = 5$, $y = 5$, $z = \lambda$:

$\frac{5 + 1}{3} = \frac{5 - 3}{1} = \frac{\lambda - 2}{4} \quad\Longrightarrow\quad 2 = 2 = \frac{\lambda - 2}{4}.$

From $\dfrac{\lambda - 2}{4} = 2$,

$\lambda - 2 = 8 \quad\Longrightarrow\quad \lambda = 10.$

Magnitude of a vector given as a cross product

Find the magnitude of $\vec A = (\hat i + 3\hat j - 2\hat k) \times (-\hat i + 3\hat k)$.

Compute the cross product as a determinant:

$\vec A = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 3 & -2 \\ -1 & 0 & 3 \end{vmatrix}.$

Expanding along the first row:

$\vec A = \hat i\,(3 \cdot 3 - (-2)\cdot 0) - \hat j\,(1 \cdot 3 - (-2)(-1)) + \hat k\,(1 \cdot 0 - 3 \cdot (-1)).$

So

$\vec A = \hat i\,(9) - \hat j\,(3 - 2) + \hat k\,(3) = 9\,\hat i - \hat j + 3\,\hat k.$

Its magnitude is

$|\vec A| = \sqrt{9^2 + (-1)^2 + 3^2} = \sqrt{81 + 1 + 9} = \sqrt{91}.$

Key takeaways

• Use the dot product to get $\cos\theta$ and the cross product magnitude to get $\sin\theta$; pick whichever the question gives you.
• An angle is obtuse exactly when the dot product is negative, which turns the problem into solving an inequality.
• For definite integrals, the symmetry property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ and a clean substitution are the two go-to tools.
• Three points are collinear when the third point satisfies the Cartesian equation of the line through the other two.