Search Previous Year Questions

Calculation:

We are given the line equation: $\sqrt{3}x + 2y = 7$

The general perpendicular form is: $x\cos\alpha + y\sin\alpha = p$

Compare with $Ax + By = C$ where $A = \sqrt{3}, B = 2, C = 7$

$\sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 2^2} = \sqrt{3 + 4} = \sqrt{7}$

Normalize coefficients

$\cos\alpha = \frac{\sqrt{3}}{\sqrt{7}}$

$\sin\alpha = \frac{2}{\sqrt{7}}$

$p = \frac{7}{\sqrt{7}} = \sqrt{7}$

Final perpendicular form

$\frac{\sqrt{3}}{\sqrt{7}}x + \frac{2}{\sqrt{7}}y = \sqrt{7}$

Therefore, the correct answer is Option 3.

Calculation:

$d = BD = \sqrt{(4 - 2)^2 + (1 - 3)^2}$

$= \sqrt{4 + 4} = 2\sqrt{2}$

Area of the square $= \frac{1}{2} d^2$

$= \frac{1}{2} (2\sqrt{2})^2 = \frac{1}{2} \times 8$

$= 4$ square units

Calculation:

$px + qy = p + q \Rightarrow m_1 = -\frac{p}{q}$

$p(x - y) + q(x + y) = 2q$

$x(p + q) - y(p - q) = 2q \Rightarrow m_2 = \frac{p + q}{p - q}$

$\tan \theta = \frac{m_2 - m_1}{1 + m_1 \cdot m_2}$

$= \left| \frac{(p + q)}{(p - q)} + \frac{p}{q} \right| \Big/ \left| 1 - \frac{p}{q} \cdot \frac{(p + q)}{(p - q)} \right|$

$= \left| \frac{pq + q^2 + p^2 - pq}{pq - q^2 - p^2 + pq} \right| = 1$

$\tan \theta = \tan\left(\frac{\pi}{4}\right) \Rightarrow \theta = \frac{\pi}{4}$

$\therefore \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

Hence, the correct answer is Option 4.

Calculation:

$x^2 + y^2 - 2kx - 2ky + k^2 = 0$

$\Rightarrow (x - k)^2 + (y - k)^2 = k^2$

$\therefore$ Centre = $(k, k)$ and radius = $k$

So circle touches the x-axis at $P(k, 0)$ and y-axis at $Q(0, k)$

$\therefore PQ = \sqrt{k^2 + k^2} = \sqrt{2k^2} = \sqrt{2}k$

Hence, the correct answer is Option 1.

Calculation:

The given equation of the hyperbola is:

$\dfrac{x^2}{1} - \dfrac{y^2}{\tfrac{1}{4}} = 1$

Comparing this with the standard form of a hyperbola

$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$

we get $a^2 = 1$ and $b^2 = \tfrac{1}{4}$.

For a hyperbola, the focal distance satisfies:

$c^2 = a^2 + b^2$

Substituting the values,

$c^2 = 1 + \tfrac{1}{4} = \tfrac{5}{4}$

Thus,

$c = \dfrac{\sqrt{5}}{2}$

The distance between the two foci of the hyperbola is:

$2c = \sqrt{5}$

Hence, the correct answer is option 2

Calculation:

$|\vec{p}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 4 + 4 - 4 = 4$

$\Rightarrow |\vec{p}| = \sqrt{4} = 2$

⇒ $|\vec{q}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 4 + 4 + 4 = 12$

$\Rightarrow |\vec{q}| = \sqrt{12} = 2\sqrt{3}$

⇒ $\vec{p} \cdot \vec{q} = |\vec{a}|^2 - |\vec{b}|^2 = 4 - 4 = 0$

$\therefore \vec{p} \perp \vec{q} \Rightarrow \theta = 90^\circ$

$|\vec{p} \times \vec{q}| = |\vec{p}||\vec{q}| \sin \theta = 2 \cdot 2\sqrt{3} \cdot \sin 90^\circ = 4\sqrt{3}$

Hence, the correct answer is Option 4.

Calculation:

A vector perpendicular to both the given vectors is obtained using the cross product.

Let the given vectors be represented in determinant form:

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

Expanding the determinant, we get:

$\vec{n} = \hat{i}(-3 - 1) - \hat{j}(6 - 1) + \hat{k}(2 + 1)$

$\vec{n} = -4\hat{i} - 5\hat{j} + 3\hat{k}$

Multiplying the vector by 2,

$2\vec{n} = -8\hat{i} - 10\hat{j} + 6\hat{k}$

Dividing by 50 to obtain the required form,

$\frac{1}{50}\vec{n} = \frac{1}{50}(-4\hat{i} - 5\hat{j} + 3\hat{k})$

Thus, Given all three vectors perpendicualr two given vectors

Hence, the correct answer is Option 4.

Calculation:

A vector perpendicular to both the given vectors is obtained using the cross product.

Let the given vectors be represented in determinant form:

⇒ $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 2 & 1 & 2 \end{vmatrix}$

Expanding the determinant, we get:

⇒ $\vec{b}_1 \times \vec{b}_2 = \hat{i}(4 - 3) - \hat{j}(2 - 6) + \hat{k}(1 - 4)$

⇒ $\vec{b}_1 \times \vec{b}_2 = \hat{i} + 4\hat{j} - 3\hat{k}$

The magnitude of the cross product (and hence the area of the parallelogram formed by the two vectors) is:

⇒ $\left\lvert \vec{b}_1 \times \vec{b}_2 \right\rvert = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$

Hence, the correct answer is option 3.

Calculation:

We calculate vector $\vec{AC}$ by subtracting the position vectors of points C and A:

$\vec{AC} = (2\hat{i} - 3\hat{j} + 2\hat{k}) - (3\hat{i} + 4\hat{j} - 2\hat{k})$

$\vec{AC} = -\hat{i} - 7\hat{j} + 4\hat{k}$

$|\vec{AC}| = \sqrt{(-1)^2 + (-7)^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}$

Next, compute vector $\vec{BD}$ using point subtraction:

$\vec{BD} = (6\hat{i} - 2\hat{j} + \hat{k}) - (4\hat{i} - 4\hat{j} - 3\hat{k}) = 2\hat{i} + 2\hat{j} + 4\hat{k}$

Now calculate the magnitude of $\vec{BD}$:

$|\vec{BD}| = \sqrt{(2)^2 + (2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24}$

Compute the dot product $\vec{AC} \cdot \vec{BD}$:

$\vec{AC} \cdot \vec{BD} = (-1)(2) + (-7)(2) + (4)(4) = -2 - 14 + 16 = 0$

Use the dot product formula to find the angle between the vectors:

$\cos \theta = \frac{\vec{AC} \cdot \vec{BD}}{|\vec{AC}| |\vec{BD}|} = \frac{0}{\sqrt{66} \cdot \sqrt{24}} = 0 \Rightarrow \theta = 90^\circ$

Hence, the correct answer is Option 1.

Calculation:

We begin by calculating the position vector $\vec{r}$ from point B to point A:

⇒ $\vec{r} = \vec{A} - \vec{B} = (1 + 1)\hat{i} + (2 + 2)\hat{j} + (5 - 3)\hat{k} = 2\hat{i} + 4\hat{j} + 2\hat{k}$

Next, compute the moment vector $\vec{M}$ using the cross product:

⇒ $\vec{M} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 2 \ 2\lambda & 2 & 5 \end{vmatrix}$

Expanding the determinant:

⇒ $\vec{M} = \hat{i}(4 \cdot 5 - 2 \cdot 2) - \hat{j}(2 \cdot 5 - 2 \cdot 2\lambda) + \hat{k}(2 \cdot 2 - 4 \cdot 2\lambda)$

⇒ $\vec{M} = \hat{i}(20 - 4) - \hat{j}(10 - 4\lambda) + \hat{k}(4 - 8\lambda) = (16 + 0)\hat{i} - 6\hat{j} + 2\hat{k}$

Equating components to solve for $\lambda$:

⇒ $16 = 20 + 2\lambda \Rightarrow \lambda = -2$

⇒ $2\lambda = -2\lambda - 8 \Rightarrow \lambda = -2$

Thus, the value of $\lambda$ is $-2$

Hence, the correct answer is Option 1.