MHT CET 2025 25th April Evening Shift (Mathematics) Question Paper with Solutions

Let the point of contact be $P(h,k)$.

Then, $ℎ+2𝑘+4=0...\left(𝑖\right)$

Slope of $𝑥+2𝑦+4=0$ is $\frac{-1}{2}$.

OP is perpendicular to $𝑥+2𝑦+4=0$.

$\begin{array}{cc}\therefore & \left(\frac{𝑘-1}{ℎ+1}\right)\left(\frac{-1}{2}\right)=-1\\ & \Rightarrow 2ℎ-𝑘=-3...\left(𝑖𝑖\right)\end{array}$

Solving (i) and (ii), we get $h=-2$ and $k=-1$

Let A: Outcome of the test was positive

$D_1$ : Patient has disease d1

$D_2$ : Patient has disease d 2

$D_3$ : Patient has disease d 3

Then, $P\left(D_1\right)=P\left(D_2\right)=P\left(D_3\right)=\frac{1}{3}$,

$\begin{array}{cc}& 𝑃(𝐴/{𝐷}_1)=0.7,𝑃(𝐴/{𝐷}_2)=0.5\\ & 𝑃(𝐴/{𝐷}_3)=0.8\\ & \text{ Required probability }=𝑃({𝐷}_2/𝐴)\\ & =\frac{𝑃\left({𝐷}_2\right)𝑃(𝐴/{𝐷}_2)}{𝑃\left({𝐷}_1\right)𝑃(𝐴/{𝐷}_1)+𝑃\left({𝐷}_2\right)𝑃(𝐴/{𝐷}_2)+𝑃\left({𝐷}_3\right)𝑃(𝐴/{𝐷}_3)}\end{array}$

$\begin{array}{cc}& =\frac{\frac{1}{3}\times 0.5}{\frac{1}{3}\times 0.7+\frac{1}{3}\times 0.5+\frac{1}{3}\times 0.8}\\ & =\frac{0.5}{2}=\frac{1}{4}\end{array}$

$\begin{array}{cc}{𝑍}_2& =(𝑟\cos (\frac{𝜋}{2}+𝜃),𝑟\sin (\frac{𝜋}{2}+𝜃))\\ & =(-𝑟\sin 𝜃,𝑟\cos 𝜃)\\ & =(-6,7)\\ & =-6+7𝑖\end{array}$

$\begin{array}{cc}& \underset{𝑥\to 0}{lim}\frac{{63}^{𝑥}-9^{𝑥}-7^{𝑥}+1}{\sqrt{\surd 2}-\sqrt{\surd 1+\cos 𝑥}}\\ & =\underset{𝑥\to 0}{lim}\frac{9^{𝑥}\cdot 7^{𝑥}-9^{𝑥}-7^{𝑥}+1}{\sqrt{\surd 2}-\sqrt{\surd 1+\cos 𝑥}}\times \frac{\sqrt{\surd 2}+\sqrt{\surd 1+\cos 𝑥}}{\sqrt{\surd 2}+\sqrt{\surd 1+\cos 𝑥}}\\ & =\underset{𝑥\to 0}{lim}\frac{[9^{𝑥}(7^{𝑥}-1)-1(7^{𝑥}-1)](\sqrt{\surd 2}+\sqrt{\surd 1+\cos 𝑥})}{2-(1+\cos 𝑥)}\\ & =\underset{𝑥\to 0}{lim}\frac{(7^{𝑥}-1)(9^{𝑥}-1)(\sqrt{\surd 2}+\sqrt{\surd 1+\cos 𝑥})}{1-\cos 𝑥}\\ & =\underset{𝑥\to 0}{lim}\frac{7^{𝑥}-1}{\frac{𝑥}{𝑥}}\cdot \frac{9^{𝑥}-1}{\frac{1-\cos 𝑥}{{𝑥}^2}}\\ & =\frac{(\log 7)(\log 9)\left(2\sqrt{\surd 2}\right)}{\frac{1}{2}}\\ & =4\sqrt{\surd 2}\log 7\log 9\end{array}$

$\begin{array}{cc}\left(𝐴\right)& 𝑝:3+2=7\\ & 𝑞:4+3=8\\ & 𝑝\to 𝑞\equiv 𝐹\to 𝐹\equiv 𝑇\end{array}$

$\therefore$ Statement (A) is true.

(B)   $p:5+2=7$

        q : Earth is flat

$p\to q\equiv T\to F\equiv F$

$\therefore$ Statement (B) is false.

$\begin{array}{cc}\left(𝐶\right)& p:\text{ Both }\left(A\right)\text{ and }\left(B\right)\\ & q:5+6=11\\ & p\to q\equiv F\to T\equiv T\end{array}$

$\therefore$ Statement $\left(C\right)$ is true.

$\begin{array}{cc}& {𝐴}_{11}=(-1{)}^{1+1}(2)=2,{𝐴}_{12}=(-1{)}^{1+2}\left(3\right)=-3\\ & {𝐴}_{21}=(-1{)}^{2+1}(-𝑏)=𝑏,{𝐴}_{22}=(-1{)}^{2+2}\left(5𝑎\right)=5𝑎\\ \therefore & \mathrm{adj}𝐴={\left[\begin{array}{cc}2& -3\\ 𝑏& 5𝑎\end{array}\right]}^{𝑇}=\left[\begin{array}{cc}2& 𝑏\\ -3& 5𝑎\end{array}\right]\end{array}$

Given, $A\mathrm{adj}A={\mathrm{AA}}^T$

$\begin{array}{cc}& \Rightarrow \left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\left[\begin{array}{cc}2& b\\ -3& 5a\end{array}\right]=\left[\begin{array}{cc}5a& -b\\ 3& 2\end{array}\right]\left[\begin{array}{cc}5a& 3\\ -b& 2\end{array}\right]\\ & \Rightarrow \left[\begin{array}{cc}10a+3b& 0\\ 0& 3b+10a\end{array}\right]=\left[\begin{array}{cc}25a^2+b^2& 15a-2b\\ 15a-2b& 13\end{array}\right]\end{array}$

$\therefore$ by the equality of matrices,

$\begin{array}{cc}& 15𝑎-2𝑏=0\text{ and }3𝑏+10𝑎=13\\ & \Rightarrow 𝑎=\frac{2}{5}\text{ and }𝑏=3\\ & \therefore 5𝑎+𝑏=5\left(\frac{2}{5}\right)+3=2+3=5\end{array}$

$\sin 𝜃=\frac{-1}{2}$ and $\tan 𝜃=\frac{1}{\sqrt{\surd 3}}$

$\Rightarrow \sin 𝜃=-\sin \frac{𝜋}{6}$ and $\tan 𝜃=\tan \frac{𝜋}{6}$

$\Rightarrow \sin 𝜃=\sin (𝜋+\frac{𝜋}{6})$ and $\tan 𝜃=\tan (𝜋+\frac{𝜋}{6})$

$\Rightarrow$ common principal solution $=𝜋+\frac{𝜋}{6}=\frac{7𝜋}{6}$

If the slopes of the lines given by $𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2=0$ are in the ratio $𝑚:𝑛$, then $(𝑚+𝑛{)}^2𝑎𝑏=4{\mathrm{mnh}}^2$.

Here, $a=6,h=h,b=1$, $m:n=2:3$

$\begin{array}{cc}\therefore & (2+3{)}^2(6\left)\right(1)=4(2\left)\right(3)h^2\\ & \Rightarrow 150=24h^2\\ & \Rightarrow h^2=\frac{25}{4}\\ & \Rightarrow h=\pm \frac{5}{2}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \sin 𝜃=\frac{|\overline{𝑎}\times \overline{𝑏}|}{\left|\overline{𝑎}\right|\left|\overline{𝑏}\right|}=\frac{5\sqrt{\surd 5}}{\left(5\right)\left(3\right)}=\frac{\sqrt{\surd 5}}{3}\\ & {\cos }^2𝜃=1-{\sin }^2𝜃=1-{\left(\frac{\sqrt{\surd 5}}{3}\right)}^2\\ \therefore & {\cos }^2𝜃=\frac{4}{9}\Rightarrow \cos 𝜃=\pm \frac{2}{3}\end{array}\\ & \text{ Since }𝜃\text{ is an obtuse angle, }\cos 𝜃<0\\ & \begin{array}{cc}\therefore \cos 𝜃& =\frac{-2}{3}\\ \overline{𝑎}\cdot \overline{𝑏}& =\left|\overline{𝑎}\right|\left|\overline{𝑏}\right|\cos 𝜃=\left(5\right)\left(3\right)\left(\frac{-2}{3}\right)=-10\end{array}\end{array}$

$\begin{array}{cc}& \text{ The vertices of triangle }\mathrm{ABC}\text{ are }\\ & \begin{array}{cc}& A\equiv (2,0,0),B\equiv (0,-3,0),C\equiv (0,0,-5)\\ \therefore & \overline{\mathrm{AB}}=-2\hat{i}-3\hat{j}\text{ and }\overline{\mathrm{AC}}=-2\hat{i}-5\hat{k}\end{array}\\ & \begin{array}{cc}\overline{\mathrm{AB}}\times \overline{\mathrm{AC}}& =\mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -2& -3& 0\\ -2& 0& -5\end{array}\mid \\ & =15\hat{i}-10\hat{j}-6\hat{k}\\ \therefore 𝐴\left(△𝐴𝐵𝐶\right)& =\frac{1}{2}|\overline{\mathrm{AB}}\times \overline{\mathrm{AC}}|\\ & =\frac{1}{2}\sqrt{\surd 225+100+36}\\ & =\frac{1}{2}\sqrt{\surd 361}=\frac{19}{2}\text{ sq. units }\end{array}\end{array}$

Let P be the population at time t years. Then,

$\begin{array}{cc}& \frac{𝑑𝑃}{𝑑𝑡}=𝑘𝑃\\ & \Rightarrow \frac{𝑑𝑃}{𝑃}=𝑘𝑑𝑡\end{array}$

Integrating on both sides, we get

$\log P=\mathrm{kt}+c$

When $t=0,P=30,000$

$\therefore \log 30,000=0+𝑐$

$\begin{array}{cc}& \Rightarrow c=\log 30,000\\ \therefore & \log P=\mathrm{kt}+\log 30,000\\ & \Rightarrow \log \left(\frac{P}{30000}\right)=\mathrm{kt}...\left(𝑖\right)\\ & \text{ When }t=40,P=40,000\\ \therefore & \log \left(\frac{40000}{30000}\right)=40k\\ & \Rightarrow k=\frac{1}{40}\log \left(\frac{4}{3}\right)\end{array}$

$\begin{array}{cc}& \begin{array}{cc}\therefore & \log \left(\frac{P}{30000}\right)=\frac{t}{40}\log \left(\frac{4}{3}\right)...\left[𝐹𝑟𝑜𝑚\right(𝑖\left)\right]\\ & \Rightarrow \log \left(\frac{P}{30000}\right)=\log {\left(\frac{4}{3}\right)}^{\frac{1}{40}}\\ & \Rightarrow P=30,000{\left(\frac{4}{3}\right)}^{\frac{1}{40}}\\ & \Rightarrow a=30,000,b=\frac{4}{3}\end{array}\\ & \end{array}$

Here, $𝑞=\frac{1}{5}$

$\therefore 𝑝=1-\frac{1}{5}=\frac{4}{5}$

Also, $n=5$

$\therefore$ Required probability $=𝑃(𝑋=4)$

$={}^5{C}_4{\left(\frac{4}{5}\right)}^4\left(\frac{1}{5}\right)=5{\left(\frac{4}{5}\right)}^4\left(\frac{1}{5}\right)={\left(\frac{4}{5}\right)}^4$

Two coins are tossed.

Sample space $=\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\}$

$\therefore n\left(S\right)=4$

$𝑃($ getting 2 heads $)=\frac{1}{4}$

$P($ getting 1 head $)=\frac{2}{4}=\frac{1}{2}$

$P($ getting no head $)=\frac{1}{4}$

Let ' X ' denote the winning amount.

$\therefore X$ can take values 10,5 and 2 .

$\therefore$ Probability distribution of X is as follows:

$\begin{array}{cccc}X=𝑥& 10& 5& 2\\ P(X=𝑥)& \frac{1}{4}& \frac{1}{2}& \frac{1}{4}\end{array}$

$\begin{array}{cc}& E\left(X\right)=\sum {𝑥}_{𝑖}P\left({𝑥}_{𝑖}\right)\\ & =10\left(\frac{1}{4}\right)+5\left(\frac{1}{2}\right)+2\left(\frac{1}{4}\right)=\frac{11}{2}\\ & \begin{array}{cc}E\left(X^2\right)& =\sum {𝑥}_2^{𝑖}P\left({𝑥}_{𝑖}\right)\\ =& 100\left(\frac{1}{4}\right)+25\left(\frac{1}{2}\right)+4\left(\frac{1}{4}\right)=\frac{77}{2}\\ \mathrm{Var}\left(X\right)& =E\left(X^2\right)-\left[E\right(X){]}^2\\ & =\frac{77}{2}-{\left(\frac{11}{2}\right)}^2\\ & =38.5-30.25\\ & =8.25\end{array}\end{array}$

$\begin{array}{cc}& \text{ Since }\sum _{𝑥=1}^5𝑃(𝑋=𝑥)=1\\ & 𝐾+2𝐾+{𝐾}^2+2𝐾+5{𝐾}^2=1\\ & \Rightarrow 6{𝐾}^2+5𝐾-1=0\\ & \Rightarrow (6𝐾-1)(𝐾+1)=0\\ & \Rightarrow 𝐾=\frac{1}{6}\dots [\because 𝐾=-1\text{ is not possible }]\\ & \begin{array}{cc}𝑃(𝑋>2)=& 1-𝑃(𝑋\le 2)\\ =& 1-(𝐾+2𝐾)\\ =& 1-3𝐾\\ =& 1-\frac{3}{6}\\ =& \frac{1}{2}\end{array}\end{array}$

$\begin{array}{cc}& (1+{𝑥}^2)\frac{𝑑𝑦}{𝑑𝑥}+2𝑥𝑦=4{𝑥}^2\\ \Rightarrow & \frac{𝑑𝑦}{𝑑𝑥}+\left(\frac{2𝑥}{1+{𝑥}^2}\right)𝑦=\frac{4{𝑥}^2}{1+{𝑥}^2}\\ \therefore & \text{ I.F. }=e^{\int \frac{2𝑥}{1+{𝑥}^2}d𝑥}=e^{\log (1+{𝑥}^2)}=1+{𝑥}^2\end{array}$

$\therefore$ Solution of the given equation is

$\begin{array}{cc}& 𝑦(1+{𝑥}^2)=\int \frac{4{𝑥}^2}{1+{𝑥}^2}\times (1+{𝑥}^2)d𝑥+c\\ & \Rightarrow 𝑦(1+{𝑥}^2)=\int 4{𝑥}^{2}d𝑥+c\\ & \Rightarrow 𝑦(1+{𝑥}^2)=\frac{4}{3}{𝑥}^3+c...\left(𝑖\right)\end{array}$

Since the curve passes through origin $(0,0)$.

$\begin{array}{cc}\therefore & 0=\frac{4}{3}\left(0\right)+𝑐\\ & \Rightarrow 𝑐=0\\ \therefore & 𝑦(1+{𝑥}^2)=\frac{4}{3}{𝑥}^3\dots \left[𝐹𝑟𝑜𝑚\right(𝑖\left)\right]\\ & \Rightarrow 3𝑦(1+{𝑥}^2)=4{𝑥}^3\end{array}$

$\begin{array}{cc}& 𝑦={\tan }^{-1}\left(\frac{12𝑥-64{𝑥}^3}{1-48{𝑥}^2}\right)\\ & \text{ Put }4𝑥=\tan 𝜃\Rightarrow 𝜃={\tan }^{-1}\left(4𝑥\right)\\ \therefore & 𝑦={\tan }^{-1}\left(\frac{3\tan 𝜃-{\tan }^3𝜃}{1-3{\tan }^2𝜃}\right)\\ & ={\tan }^{-1}(\tan 3𝜃)\\ & =3𝜃\\ \therefore & 𝑦=3{\tan }^{-1}\left(4𝑥\right)\\ \therefore & \frac{d𝑦}{d𝑥}=\frac{3}{1+(4𝑥{)}^2}\times 4=\frac{12}{1+16{𝑥}^2}\end{array}$

$\begin{array}{cc}𝑦& =({𝐶}_1+{𝐶}_2)\sin (𝑥+{𝐶}_3)-{𝐶}_4{e}^{𝑥+{𝐶}_5}\\ & =({𝐶}_1+{𝐶}_2)\sin (𝑥+{𝐶}_3)-{𝐶}_4{e}^{𝑥}\cdot e^{{𝐶}_5}\\ & =𝐴\sin (𝑥+{𝐶}_3)-𝐵e^{𝑥}\\ & \dots [𝐴={𝐶}_1+{𝐶}_2,𝐵={𝐶}_4{e}^{{𝐶}_5}]\end{array}$

There are 3 arbitrary constants.

$\therefore$     Order of the differential equation is 3 .

The feasible region lies on the origin side of $𝑥+𝑦=5$ and $4𝑥+𝑦=12$, on the non-origin side of $𝑥+2𝑦=4$, and it is in the first quadrant.

The corner points of the feasible region are

$\begin{array}{cc}& A(0,2),B(\frac{20}{7},\frac{4}{7}),C(\frac{7}{3},\frac{8}{3})\text{ and }D(0,5)\\ & \text{ At }A(0,2),Z=6\left(0\right)+3\left(2\right)=6\\ & \text{ At }B(\frac{20}{7},\frac{4}{7}),Z=6\left(\frac{20}{7}\right)+3\left(\frac{4}{7}\right)=\frac{132}{7}\\ & \text{ At }C(\frac{7}{3},\frac{8}{3}),Z=6\left(\frac{7}{3}\right)+3\left(\frac{8}{3}\right)=22\\ & \text{ At }D(0,5)=6\left(0\right)+3\left(5\right)=15\end{array}$

$\therefore$ Maximum value of Z is 22 .

The given lines can be written as

$\frac{𝑥}{1}=\frac{𝑦-\frac{1}{a}}{\frac{1}{a}}=\frac{𝑧-2}{1}$ and

$\frac{𝑥}{1}=\frac{𝑦-\frac{2}{3}}{\frac{1}{3}}=\frac{𝑧-\frac{2}{𝑏}}{\frac{1}{𝑏}}$

Since the lines are coplanar.

$\begin{array}{cc}\therefore & \mid \begin{array}{ccc}0& \frac{2}{3}-\frac{1}{𝑎}& \frac{2}{𝑏}-2\\ 1& \frac{1}{𝑎}& 1\\ 1& \frac{1}{3}& \frac{1}{𝑏}\end{array}\mid =0\\ & \Rightarrow -(\frac{2}{3}-\frac{1}{𝑎})(\frac{1}{𝑏}-1)+(\frac{2}{𝑏}-2)(\frac{1}{3}-\frac{1}{𝑎})=0\\ & \Rightarrow \frac{1}{𝑎𝑏}-\frac{1}{𝑎}=0\\ & \Rightarrow 𝑏=1,𝑎\in 𝑅-\left\{0\right\}\dots [𝑎𝑏\ne 0]\end{array}$

$\begin{array}{cc}& {({\tan }^{-1}𝑥)}^2+{({\cot }^{-1}𝑥)}^2=\frac{5{𝜋}^2}{8}\\ & \Rightarrow {({\tan }^{-1}𝑥+{\cot }^{-1}𝑥)}^2-2{\tan }^{-1}𝑥{\cot }^{-1}𝑥=\frac{5{𝜋}^2}{8}\\ & \Rightarrow {({\tan }^{-1}𝑥+{\cot }^{-1}𝑥)}^2-2{\tan }^{-1}𝑥(\frac{𝜋}{2}-{\tan }^{-1}𝑥)=\frac{5{𝜋}^2}{8}\end{array}$

$\begin{array}{cc}& \Rightarrow \frac{{𝜋}^2}{4}-2\times \frac{𝜋}{2}{\tan }^{-1}𝑥+2{({\tan }^{-1}𝑥)}^2=\frac{5{𝜋}^2}{8}\\ & \Rightarrow 2{({\tan }^{-1}𝑥)}^2-𝜋{\tan }^{-1}𝑥-\frac{3{𝜋}^2}{8}=0\\ & \Rightarrow {\tan }^{-1}𝑥=-\frac{𝜋}{4},\frac{3𝜋}{4}\\ & \Rightarrow {\tan }^{-1}𝑥=-\frac{𝜋}{4}\\ & \Rightarrow 𝑥=-1\\ & \Rightarrow {𝑥}^2+1=2\end{array}$

The symbolic form is

$\begin{array}{cc}& [𝑝\wedge (𝑞\vee 𝑟\left)\right]\vee (\sim 𝑟\wedge \sim 𝑞\wedge 𝑝)\\ & \equiv [𝑝\wedge (𝑞\vee 𝑟\left)\right]\vee \left[\right(\sim 𝑟\wedge \sim 𝑞)\wedge 𝑝]\dots \left[𝐴𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑣𝑒𝑙𝑎𝑤\right]\end{array}$

$\begin{array}{cc}& \equiv [𝑝\wedge (𝑞\vee 𝑟\left)\right]\vee [𝑝\wedge (\sim 𝑞\wedge \sim 𝑟\left)\right]\text{ ...[Commutative law] }\end{array}$

$\equiv 𝑝\wedge \left[\right(𝑞\vee 𝑟)\vee (\sim 𝑞\wedge \sim 𝑟\left)\right]\dots \left[𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒𝑙𝑎𝑤\right]$

$\begin{array}{cc}& \equiv 𝑝\wedge \left[\right(𝑞\vee 𝑟)\vee \sim (𝑞\vee 𝑟\left)\right]\text{ ...[De Morgan’s law] }\end{array}$

$\equiv p\wedge T>\dots \left[𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑙𝑎𝑤\right]$

$\equiv p>\dots \left[𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦𝑙𝑎𝑤\right]$

$\begin{array}{cc}& \text{ Since }g\left(𝑥\right)\text{ is continuous at }𝑥=3\text{. }\\ & \begin{array}{cc}\therefore & 𝑔\left(3\right)=\underset{𝑥\to 3}{lim}{\int }_{r\left(𝑥\right)}^3\frac{3t^2}{𝑥-3}\mathrm{dt}\\ \Rightarrow & 𝐾=\underset{𝑥\to 3}{lim}\frac{{\int }_{f\left(𝑥\right)}^{3}3t^2\mathrm{dt}}{𝑥-3}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Applying L’Hospital’s rule, we get }\\ & \begin{array}{cc}𝐾& =\underset{𝑥\to 3}{lim}\frac{3\left[𝑓\right(𝑥\left){]}^2{𝑓}^{\prime }\right(𝑥)}{1}\\ & =3\left[𝑓\right(3\left){]}^2{𝑓}^{\prime }\right(3)=3(3{)}^2\left(\frac{1}{27}\right)=1\end{array}\end{array}$

$\begin{array}{cc}\text{ Required area }& ={\int }_4^0(4𝑥-{𝑥}^2)d𝑥\\ & ={[\frac{4{𝑥}^2}{2}-\frac{{𝑥}^3}{3}]}_4^0=32-\frac{64}{3}\\ & =\frac{32}{3}\text{ sq. units }\end{array}$

The given plane passes through the point $(2,-3,0)$ and perpendicular to the vectors $\bar{p}=\hat{i}+2\hat{j}-\hat{k}$ and $\bar{q}=2\hat{i}+3\hat{j}+\hat{k}$

Here, $\overline{𝑎}=2\widehat{𝑖}-3\widehat{𝑗}$

$\begin{array}{cc}& \overline{𝑛}=\overline{𝑝}\times \overline{𝑞}=\mid \begin{array}{ccc}\widehat{𝑖}& \widehat{𝑗}& \widehat{𝑘}\\ 1& 2& -1\\ 2& 3& 1\end{array}\mid \\ & =\widehat{𝑖}(2+3)-\widehat{𝑗}(1+2)+\widehat{𝑘}(3-4)\\ & =5\widehat{𝑖}-3\widehat{𝑗}-\widehat{𝑘}\end{array}$