JEE Main 2025 — Mathematics Vectors & 3D Geometry

Line is $\perp^{\mathrm{r}}$⊥r to 2 line $\Rightarrow$⇒ line will be parallel to

$$(i+a \hat{j}+b \hat{k}) \times(-b \hat{i}+a \hat{j}+5 \hat{k})$$(i+aj^​+bk^)×(−bi^+aj^​+5k^)

Parallel vector along the required line is

$$\hat{\mathrm{i}}(5 \mathrm{a}-\mathrm{ab})-\hat{\mathrm{j}}\left(\mathrm{~b}^2+5\right)+\hat{\mathrm{k}}(\mathrm{a}+\mathrm{ab})$$i^(5a−ab)−j^​( b2+5)+k^(a+ab)

Dr's of required line $\alpha(5 a-a b),-\left(b^2+5\right),(a+a b)$α(5a−ab),−(b2+5),(a+ab)

Also Dr's of required line $\alpha-2, \mathrm{~d},-4$α−2, d,−4

Also point $\left(0, \frac{-1}{2}, 0\right)$(0,2−1​,0) will lie on $\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}+4}{\mathrm{~d}}=\frac{\mathrm{z}-\mathrm{c}}{-4}$−2x−1​= dy+4​=−4z−c​

$$\frac{0-1}{-2}=\frac{\frac{-1}{2}+4}{d}=\frac{0-c}{-4} \Rightarrow d=7, c=2$$−20−1​=d2−1​+4​=−40−c​⇒d=7,c=2

$\begin{aligned}
&\text { From (1) } \frac{5 \mathrm{a}-\mathrm{ab}}{-2}=\frac{-\mathrm{b}^2-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}\
&\frac{5 \mathrm{a}-\mathrm{ab}}{-2}=\frac{\mathrm{a}+\mathrm{ab}}{-4} ; \frac{-\mathrm{b}^2-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}
\end{aligned}$

\begin{array}{c|c}-20 a+4 a b=-2 a-2 a b & 4 b^2+20=70+7 a b \18 a=6 a b & 36+20=70+21 a \b=3 & 56=28 a \Rightarrow a=2\end{array}

$$a+b+c+d=2+3+2+7=14$$a+b+c+d=2+3+2+7=14