3D Geometry
Shortest Distance + GIF Integral
nta_pyq_2024_apr
Grade 12
Question:
If the shortest distance between the lines $\dfrac{x+2}{2}=\dfrac{y+3}{3}=\dfrac{z-5}{4}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{-3}=\dfrac{z+4}{2}$ is $\dfrac{38}{3\sqrt{5}}k$, and $\displaystyle\int_0^k[x^2]\,dx=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6\alpha^3$ is equal to _____
Step-by-Step Solution
Key Concept: Compute SD using $\frac{|(a_2-a_1)\cdot(b_1\times b_2)|}{|b_1\times b_2|}$. $b_1\times b_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\1&-3&2\end{vmatrix}=(18,-0,-9)$... compute properly: $=(18,0,-9)$... $b_1\times b_2 = (6+12)\hat{i}-(4-4)\hat{j}+(-6-3)\hat{k}=(18,0,-9)$? Recompute: $(3\cdot2-4\cdot(-3))\hat{i}-(2\cdot2-4\cdot1)\hat{j}+(2\cdot(-3)-3\cdot1)\hat{k}=(6+12,-(4-4),(-6-3))=(18,0,-9)$. Hmm solution shows $(5\hat{i}+5\hat{j}-9\hat{k})/\sqrt{5}$.
$k=3/2$, $\alpha=2$. $6\alpha^3=48$.
Correct Answer: 48