MHT CET 2024 Mathematics Question Paper with Answer and Solution

(A) Let the scalar triple product be denoted by $[\bar{a} \bar{b} \bar{c}] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
We need to evaluate $(\bar{u}+\bar{v}-\bar{w}) \cdot [(\bar{u}-\bar{v}) \times (\bar{v}-\bar{w})]$.
First,expand the cross product: $(\bar{u}-\bar{v}) \times (\bar{v}-\bar{w}) = \bar{u} \times \bar{v} - \bar{u} \times \bar{w} - \bar{v} \times \bar{v} + \bar{v} \times \bar{w}$.
Since $\bar{v} \times \bar{v} = 0$,this simplifies to $\bar{u} \times \bar{v} - \bar{u} \times \bar{w} + \bar{v} \times \bar{w}$.
Now,take the dot product with $(\bar{u}+\bar{v}-\bar{w})$:
$(\bar{u}+\bar{v}-\bar{w}) \cdot (\bar{u} \times \bar{v} - \bar{u} \times \bar{w} + \bar{v} \times \bar{w})$
$= \bar{u} \cdot (\bar{u} \times \bar{v}) - \bar{u} \cdot (\bar{u} \times \bar{w}) + \bar{u} \cdot (\bar{v} \times \bar{w}) + \bar{v} \cdot (\bar{u} \times \bar{v}) - \bar{v} \cdot (\bar{u} \times \bar{w}) + \bar{v} \cdot (\bar{v} \times \bar{w}) - \bar{w} \cdot (\bar{u} \times \bar{v}) + \bar{w} \cdot (\bar{u} \times \bar{w}) - \bar{w} \cdot (\bar{v} \times \bar{w})$.
Using the property that the scalar triple product is zero if any two vectors are identical:
$= 0 - 0 + [\bar{u} \bar{v} \bar{w}] + 0 - [\bar{v} \bar{u} \bar{w}] + 0 - [\bar{w} \bar{u} \bar{v}] + 0 - 0$.
$= [\bar{u} \bar{v} \bar{w}] + [\bar{u} \bar{v} \bar{w}] - [\bar{u} \bar{v} \bar{w}] = [\bar{u} \bar{v} \bar{w}] = \bar{u} \cdot (\bar{v} \times \bar{w})$.

(C) We know that the scalar triple product is defined as $[\bar{x} \quad \bar{y} \quad \bar{z}] = (\bar{x} \times \bar{y}) \cdot \bar{z}$.
Given expression: $[\bar{a} \times \bar{b} \quad \bar{b} \times \bar{c} \quad \bar{c} \times \bar{a}] = ((\bar{a} \times \bar{b}) \times (\bar{b} \times \bar{c})) \cdot (\bar{c} \times \bar{a})$.
Using the vector identity $(\bar{u} \times \bar{v}) \times \bar{w} = (\bar{u} \cdot \bar{w})\bar{v} - (\bar{v} \cdot \bar{w})\bar{u}$,we have:
$(\bar{a} \times \bar{b}) \times (\bar{b} \times \bar{c}) = [\bar{a} \quad \bar{b} \quad \bar{c}]\bar{b}$.
Substituting this into the expression:
$[\bar{a} \times \bar{b} \quad \bar{b} \times \bar{c} \quad \bar{c} \times \bar{a}] = ([\bar{a} \quad \bar{b} \quad \bar{c}]\bar{b}) \cdot (\bar{c} \times \bar{a})$
$= [\bar{a} \quad \bar{b} \quad \bar{c}] (\bar{b} \cdot (\bar{c} \times \bar{a}))$
$= [\bar{a} \quad \bar{b} \quad \bar{c}] [\bar{b} \quad \bar{c} \quad \bar{a}]$
Since the scalar triple product is cyclic,$[\bar{b} \quad \bar{c} \quad \bar{a}] = [\bar{a} \quad \bar{b} \quad \bar{c}]$.
Therefore,$[\bar{a} \times \bar{b} \quad \bar{b} \times \bar{c} \quad \bar{c} \times \bar{a}] = [\bar{a} \quad \bar{b} \quad \bar{c}]^2$.
Comparing this with $\lambda [\bar{a} \quad \bar{b} \quad \bar{c}]^2$,we get $\lambda = 1$.

(A) Since $\overline{a}, \overline{b}, \overline{c}$ are coplanar vectors,their scalar triple product is zero,i.e.,$[\overline{a}, \overline{b}, \overline{c}] = 0$.
Let $\overline{\alpha} = 2 \overline{a} - \overline{b}$,$\overline{\beta} = 2 \overline{b} - \overline{c}$,and $\overline{\gamma} = 2 \overline{c} - \overline{a}$.
The scalar triple product $[\overline{\alpha}, \overline{\beta}, \overline{\gamma}]$ can be expressed as the determinant of the coefficients multiplied by $[\overline{a}, \overline{b}, \overline{c}]$:
$[\overline{\alpha}, \overline{\beta}, \overline{\gamma}] = \begin{vmatrix} 2 & -1 & 0 \\ 0 & 2 & -1 \\ -1 & 0 & 2 \end{vmatrix} [\overline{a}, \overline{b}, \overline{c}]$.
Calculating the determinant: $2(4 - 0) - (-1)(0 - 1) + 0 = 2(4) + 1(-1) = 8 - 1 = 7$.
Thus,$[\overline{\alpha}, \overline{\beta}, \overline{\gamma}] = 7 \times [\overline{a}, \overline{b}, \overline{c}] = 7 \times 0 = 0$.

(B) First,note that the determinant on the right side is equal to the scalar triple product $[\bar{a} \quad \bar{b} \quad \bar{c}]$.
$\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = [\bar{a} \quad \bar{b} \quad \bar{c}]$.
Now,calculate the scalar triple product on the left side:
$[3 \bar{a}+\bar{b} \quad 3 \bar{b}+\bar{c} \quad 3 \bar{c}+\bar{a}] = (3 \bar{a}+\bar{b}) \cdot ((3 \bar{b}+\bar{c}) \times (3 \bar{c}+\bar{a}))$.
Expanding the cross product:
$(3 \bar{b}+\bar{c}) \times (3 \bar{c}+\bar{a}) = 9(\bar{b} \times \bar{c}) + 3(\bar{b} \times \bar{a}) + 3(\bar{c} \times \bar{c}) + (\bar{c} \times \bar{a}) = 9(\bar{b} \times \bar{c}) + 3(\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a})$.
Now take the dot product with $(3 \bar{a}+\bar{b})$:
$= (3 \bar{a}+\bar{b}) \cdot (9(\bar{b} \times \bar{c}) + 3(\bar{b} \times \bar{a}) + (\bar{c} \times \bar{a}))$
$= 27[\bar{a} \quad \bar{b} \quad \bar{c}] + 9[\bar{a} \quad \bar{b} \quad \bar{a}] + 3[\bar{a} \quad \bar{c} \quad \bar{a}] + 9[\bar{b} \quad \bar{b} \quad \bar{c}] + 3[\bar{b} \quad \bar{b} \quad \bar{a}] + [\bar{b} \quad \bar{c} \quad \bar{a}]$.
Since any scalar triple product with two identical vectors is $0$,this simplifies to:
$= 27[\bar{a} \quad \bar{b} \quad \bar{c}] + [\bar{b} \quad \bar{c} \quad \bar{a}] = 27[\bar{a} \quad \bar{b} \quad \bar{c}] + [\bar{a} \quad \bar{b} \quad \bar{c}] = 28[\bar{a} \quad \bar{b} \quad \bar{c}]$.
Comparing this with $\lambda [\bar{a} \quad \bar{b} \quad \bar{c}]$,we get $\lambda = 28$.

(D) The scalar triple product is defined as $\left[\bar{a} \bar{b} \bar{c}\right] = \bar{a} \cdot (\bar{b} \times \bar{c})$.
Since $\bar{a}$ is perpendicular to both $\bar{b}$ and $\bar{c}$,$\bar{a}$ must be parallel to the vector $\bar{b} \times \bar{c}$.
Let $\hat{n}$ be the unit vector perpendicular to $\bar{b}$ and $\bar{c}$. Then $\bar{b} \times \bar{c} = |\bar{b}||\bar{c}| \sin\left(\frac{\pi}{3}\right) \hat{n} = 3 \times 4 \times \frac{\sqrt{3}}{2} \hat{n} = 6\sqrt{3} \hat{n}$.
Since $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$\bar{a} = |\bar{a}| \hat{n} = 2 \hat{n}$.
Therefore,$\left[\bar{a} \bar{b} \bar{c}\right] = \bar{a} \cdot (\bar{b} \times \bar{c}) = (2 \hat{n}) \cdot (6\sqrt{3} \hat{n}) = 12\sqrt{3} (\hat{n} \cdot \hat{n}) = 12\sqrt{3} \times 1 = 12\sqrt{3}$.

(A) Given the vector triple product identity: $\hat{a} \times (\hat{b} \times \hat{c}) = (\hat{a} \cdot \hat{c}) \hat{b} - (\hat{a} \cdot \hat{b}) \hat{c}$.
Comparing this with the given equation: $(\hat{a} \cdot \hat{c}) \hat{b} - (\hat{a} \cdot \hat{b}) \hat{c} = \frac{\sqrt{3}}{2} \hat{b} + \frac{\sqrt{3}}{2} \hat{c}$.
Since $\hat{b}$ and $\hat{c}$ are not parallel,we can equate the coefficients of $\hat{b}$ and $\hat{c}$:
$\hat{a} \cdot \hat{c} = \frac{\sqrt{3}}{2}$ and $-(\hat{a} \cdot \hat{b}) = \frac{\sqrt{3}}{2} \Rightarrow \hat{a} \cdot \hat{b} = -\frac{\sqrt{3}}{2}$.
Let $\theta$ be the angle between $\hat{a}$ and $\hat{b}$. Since $\hat{a}$ and $\hat{b}$ are unit vectors,$\hat{a} \cdot \hat{b} = |\hat{a}| |\hat{b}| \cos \theta = \cos \theta$.
Therefore,$\cos \theta = -\frac{\sqrt{3}}{2}$.
Since $\cos \theta = -\frac{\sqrt{3}}{2}$,we have $\theta = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$.

(D) Using the vector triple product identity $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we evaluate each term:
$\hat{i} \times (\vec{a} \times \hat{i}) = (\hat{i} \cdot \hat{i})\vec{a} - (\hat{i} \cdot \vec{a})\hat{i} = 1(\vec{a}) - (1)\hat{i} = (\hat{i} + 2\hat{j} + 3\hat{k}) - \hat{i} = 2\hat{j} + 3\hat{k}$
$\hat{j} \times (\vec{a} \times \hat{j}) = (\hat{j} \cdot \hat{j})\vec{a} - (\hat{j} \cdot \vec{a})\hat{j} = 1(\vec{a}) - (2)\hat{j} = (\hat{i} + 2\hat{j} + 3\hat{k}) - 2\hat{j} = \hat{i} + 3\hat{k}$
$\hat{k} \times (\vec{a} \times \hat{k}) = (\hat{k} \cdot \hat{k})\vec{a} - (\hat{k} \cdot \vec{a})\hat{k} = 1(\vec{a}) - (3)\hat{k} = (\hat{i} + 2\hat{j} + 3\hat{k}) - 3\hat{k} = \hat{i} + 2\hat{j}$
Summing these results:
$\vec{b} = (2\hat{j} + 3\hat{k}) + (\hat{i} + 3\hat{k}) + (\hat{i} + 2\hat{j}) = 2\hat{i} + 4\hat{j} + 6\hat{k}$
Finally,the magnitude is:
$|\vec{b}| = \sqrt{2^2 + 4^2 + 6^2} = \sqrt{4 + 16 + 36} = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$

(C) First,we check the magnitudes and dot product of $\overline{a}$ and $\overline{b}$.
$|\overline{a}|^2 = \frac{1}{10}(16+9+1) = \frac{26}{10} = 2.6$ (Note: The provided vector $\overline{a}$ is not a unit vector. Let us proceed with the given vectors).
$\overline{a} \cdot \overline{b} = \frac{1}{3\sqrt{10}}(4-6+2) = 0$.
Since $\overline{a} \cdot \overline{b} = 0$,the vectors are perpendicular.
Let $E = (2 \bar{a}-\bar{b}) \cdot \{(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b})\}$.
Using the vector triple product identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,we have:
$(\bar{a} \times \bar{b}) \times (\bar{a}+2 \bar{b}) = (\bar{a} \cdot (\bar{a}+2 \bar{b}))\bar{b} - (\bar{b} \cdot (\bar{a}+2 \bar{b}))\bar{a}$.
Since $\bar{a} \cdot \bar{b} = 0$,this simplifies to:
$= (\bar{a} \cdot \bar{a})\bar{b} - (2\bar{b} \cdot \bar{b})\bar{a} = |\bar{a}|^2 \bar{b} - 2|\bar{b}|^2 \bar{a}$.
Now,$E = (2 \bar{a}-\bar{b}) \cdot (|\bar{a}|^2 \bar{b} - 2|\bar{b}|^2 \bar{a})$.
$E = 2|\bar{a}|^2 (\bar{a} \cdot \bar{b}) - 4|\bar{a}|^2 |\bar{b}|^2 - |\bar{a}|^2 (\bar{b} \cdot \bar{b}) + 2|\bar{b}|^2 (\bar{b} \cdot \bar{a})$.
Since $\bar{a} \cdot \bar{b} = 0$,$E = -4|\bar{a}|^2 |\bar{b}|^2 - |\bar{a}|^2 |\bar{b}|^2 = -5|\bar{a}|^2 |\bar{b}|^2$.
Given $|\bar{a}|^2 = 2.6$ and $|\bar{b}|^2 = \frac{1}{9}(1+4+4) = 1$.
$E = -5(2.6)(1) = -13$. (Note: If the problem assumes unit vectors,the result is $-5$).

(A) Using the vector triple product formula,$\bar{a} \times(\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c}$.
Given $\bar{a} \times(\bar{b} \times \bar{c}) = \frac{\bar{b}+\bar{c}}{\sqrt{2}}$,we have:
$(\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c} = \frac{1}{\sqrt{2}} \bar{b} + \frac{1}{\sqrt{2}} \bar{c}$.
Rearranging the terms,we get:
$(\bar{a} \cdot \bar{c} - \frac{1}{\sqrt{2}}) \bar{b} - (\bar{a} \cdot \bar{b} + \frac{1}{\sqrt{2}}) \bar{c} = 0$.
Since $\bar{b}$ and $\bar{c}$ are non-coplanar (and thus linearly independent),the coefficients must be zero:
$\bar{a} \cdot \bar{b} + \frac{1}{\sqrt{2}} = 0 \Rightarrow \bar{a} \cdot \bar{b} = -\frac{1}{\sqrt{2}}$.
Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$.
Thus,$|\bar{a}| |\bar{b}| \cos \theta = -\frac{1}{\sqrt{2}}$,where $\theta$ is the angle between $\bar{a}$ and $\bar{b}$.
$\cos \theta = -\frac{1}{\sqrt{2}}$.
Therefore,$\theta = \frac{3 \pi}{4}$.

(D) Given the vector triple product formula: $\bar{a} \times (\bar{b} \times \bar{c}) = (\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c}$.
Comparing this with the given equation $\bar{a} \times (\bar{b} \times \bar{c}) = \frac{\sqrt{3}}{2} \bar{b} + \frac{\sqrt{3}}{2} \bar{c}$,we have:
$(\bar{a} \cdot \bar{c}) \bar{b} - (\bar{a} \cdot \bar{b}) \bar{c} = \frac{\sqrt{3}}{2} \bar{b} + \frac{\sqrt{3}}{2} \bar{c}$.
Since $\bar{b}$ and $\bar{c}$ are not parallel,we can equate the coefficients of $\bar{b}$ and $\bar{c}$:
$\bar{a} \cdot \bar{c} = \frac{\sqrt{3}}{2}$ and $-(\bar{a} \cdot \bar{b}) = \frac{\sqrt{3}}{2}$,which implies $\bar{a} \cdot \bar{b} = -\frac{\sqrt{3}}{2}$.
Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$. Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$.
$\bar{a} \cdot \bar{b} = |\bar{a}| |\bar{b}| \cos \theta = 1 \times 1 \times \cos \theta = \cos \theta$.
Thus,$\cos \theta = -\frac{\sqrt{3}}{2}$.
Since $\cos \theta = -\frac{\sqrt{3}}{2}$,the angle $\theta = \frac{5 \pi}{6}$.

(A) First,calculate the cross product $\overline{a} \times \overline{b}$:
$\overline{a} \times \overline{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-2)) + \hat{k}(2 - 1) = 2\hat{i} - 2\hat{j} + \hat{k}$.
The magnitude is $|\overline{a} \times \overline{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = 3$.
Given $|\overline{c}-\overline{a}|=2 \sqrt{2}$,squaring both sides gives $|\overline{c}|^2 + |\overline{a}|^2 - 2(\overline{c} \cdot \overline{a}) = 8$.
Since $|\overline{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = 3$ and $\overline{a} \cdot \overline{c} = |\overline{c}|$,we have $|\overline{c}|^2 + 9 - 2|\overline{c}| = 8$.
$|\overline{c}|^2 - 2|\overline{c}| + 1 = 0 \Rightarrow (|\overline{c}| - 1)^2 = 0 \Rightarrow |\overline{c}| = 1$.
The magnitude of the cross product is $|(\overline{a} \times \overline{b}) \times \overline{c}| = |\overline{a} \times \overline{b}| |\overline{c}| \sin(30^{\circ})$.
Substituting the values: $3 \times 1 \times \frac{1}{2} = \frac{3}{2}$.

(C) Given,$(\overline{a} \times \overline{b}) \times \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$.
We know the vector triple product formula: $(\overline{a} \times \overline{b}) \times \overline{c} = (\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a}$.
Comparing the two expressions,we have:
$(\overline{a} \cdot \overline{c}) \overline{b} - (\overline{b} \cdot \overline{c}) \overline{a} = \frac{1}{3}|\overline{b}||\overline{c}| \overline{a}$.
Since $\overline{a}, \overline{b}, \overline{c}$ are non-zero and no two are collinear,$\overline{a}$ and $\overline{b}$ are linearly independent. Thus,the coefficient of $\overline{b}$ must be zero:
$\overline{a} \cdot \overline{c} = 0$.
Equating the coefficients of $\overline{a}$:
$-\overline{b} \cdot \overline{c} = \frac{1}{3}|\overline{b}||\overline{c}|$.
Using the definition of the dot product $\overline{b} \cdot \overline{c} = |\overline{b}||\overline{c}| \cos \theta$:
$-|\overline{b}||\overline{c}| \cos \theta = \frac{1}{3}|\overline{b}||\overline{c}|$.
Since the vectors are non-zero,we can divide by $|\overline{b}||\overline{c}|$:
$\cos \theta = -\frac{1}{3}$.
Now,using the identity $\sin^2 \theta + \cos^2 \theta = 1$:
$\sin^2 \theta = 1 - (-\frac{1}{3})^2 = 1 - \frac{1}{9} = \frac{8}{9}$.
Since $\theta$ is the angle between two vectors,$0 \le \theta \le \pi$,so $\sin \theta \ge 0$:
$\sin \theta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

(A) Given,$\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2} \quad ...(i)$
We know that $\operatorname{cosec}^{-1}(z) = \sin^{-1}(\frac{1}{z})$.
Therefore,$\operatorname{cosec}^{-1}\left(\frac{13}{12}\right) = \sin^{-1}\left(\frac{12}{13}\right)$.
Substituting this into equation $(i)$,we get:
$\sin ^{-1}\left(\frac{x}{13}\right)+\sin ^{-1}\left(\frac{12}{13}\right)=\frac{\pi}{2}$.
We know the identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$.
Also,$\sin^{-1}(\frac{12}{13}) = \cos^{-1}(\sqrt{1 - (\frac{12}{13})^2}) = \cos^{-1}(\sqrt{1 - \frac{144}{169}}) = \cos^{-1}(\sqrt{\frac{25}{169}}) = \cos^{-1}(\frac{5}{13})$.
So,$\sin^{-1}(\frac{x}{13}) + \cos^{-1}(\frac{5}{13}) = \frac{\pi}{2}$.
Comparing this with $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$,we must have $\frac{x}{13} = \frac{5}{13}$.
Thus,$x = 5$.