If the shortest distance between the lines \fracx+22=\fracy+33=\fracz-54 and \fracx-31=\fracy-2-3=\fracz+42 is \frac383 \sqrt5 k , and \int\limits_0^k[x^2] d x=\alpha-\sqrt\alpha , where [x] denotes the greatest integer function, then 6 \alpha^3 is equal to \underline\hspace2cm .

Definite Integration PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q52 with Solution

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Correct answer: 48

The lines are $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}.$

So we can identify:

And:

Shortest distance between two lines is $D=\frac{|(\vec{B}-\vec{A})\cdot(\vec{d_1}\times \vec{d_2})|}{|\vec{d_1}\times \vec{d_2}|}.$

First, $\vec{B}-\vec{A}=(3-(-2),\,2-(-3),\,-4-5)=(5,5,-9).$

Now compute cross product:

\begin{vmatrix} \hat i & \hat j & \hat k\\ 2&3&4\\ 1&-3&2 \end{vmatrix}.$$ So, $$\vec{d_1}\times \vec{d_2}= \hat i(3\cdot 2-4\cdot(-3))- \hat j(2\cdot 2-4\cdot 1)+ \hat k(2\cdot(-3)-3\cdot 1).$$ $$=\hat i(6+12)-\hat j(4-4)+\hat k(-6-3) =(18,0,-9).$$ Hence, $$|\vec{d_1}\times \vec{d_2}|=\sqrt{18^2+0^2+(-9)^2}=\sqrt{324+81}=\sqrt{405}=9\sqrt{5}.$$ Now, $$(\vec{B}-\vec{A})\cdot(\vec{d_1}\times \vec{d_2})=(5,5,-9)\cdot(18,0,-9)=90+0+81=171.$$ Therefore, $$D=\frac{171}{9\sqrt{5}}=\frac{19}{\sqrt{5}}=\frac{19\sqrt{5}}{5}.$$ Given that $$D=\frac{38}{3\sqrt{5}}\,k.$$ So, $$\frac{19}{\sqrt{5}}=\frac{38}{3\sqrt{5}}k.$$ Multiplying by $\sqrt{5}$: $$19=\frac{38}{3}k$$ $$k=\frac{19\cdot 3}{38}=\frac{3}{2}.$$ --- 3. **Evaluate** $\int_0^k [x^2]dx$ Since $k=\frac32$, $$I=\int_0^{3/2}[x^2]dx.$$ Now $[x^2]$ changes value when $x^2$ crosses integers. For $x\ge 0$: - $[x^2]=0$ for $0\le x<1$ - $[x^2]=1$ for $1\le x<\sqrt2$ - $[x^2]=2$ for $\sqrt2\le x<\sqrt3$ But $\frac32<\sqrt3$, so only these intervals matter. Thus, $$I=\int_0^1 0\,dx+\int_1^{\sqrt2}1\,dx+\int_{\sqrt2}^{3/2}2\,dx.$$ $$I=0+(\sqrt2-1)+2\left(\frac32-\sqrt2\right).$$ $$I=\sqrt2-1+3-2\sqrt2=2-\sqrt2.$$ Given $$\int_0^k [x^2]dx=\alpha-\sqrt{\alpha}.$$ So, $$\alpha-\sqrt{\alpha}=2-\sqrt2.$$ This matches for $$\alpha=2,$$ since $$2-\sqrt2=2-\sqrt2.$$ --- 4. **Find** $6\alpha^3$ $$6\alpha^3=6(2^3)=6\cdot 8=48.$$ --- 5. **Comparison with stored answer** Derived answer = $48$ Stored correct answer = $48$ So the answer agrees.