JEE Main 2026 — Mathematics Vectors & 3D Geometry
Let the lines L1:r=i^+2j^+3k^+λ(2i^+3j^+4k^),λ∈R and L2:r=(4i^+j^)+μ(5i^+2j^+k^),μ∈R, intersect at the point R. Let P and Q be the points lying on lines L1 and L2, respectively, such that ∣PR∣=29 and ∣PQ∣=347. If the point P lies in the first octant, then 27(QR)2 is equal to
Held on 24 Jan 2026 · Verified 6 Jul 2026.
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If the distance of the point $\mathrm{P}(43, \alpha, \beta), \beta<0$, from the line $\overrightarrow{\mathrm{r}}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in \mathbf{R}$ along a line with direction ratios $3,-1,0$ is $13 \sqrt{10}$, then $\alpha^{2}+\beta^{2}$ is equal to $\_\_\_\_$
The volume of the parallelepiped formed by vectors a=i+2j-k, b=2i-j+3k, c=3i+j+2k is:
If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to
If the point of intersection of the lines $\dfrac{x+1}{3} = \dfrac{y+a}{5} = \dfrac{z+b+1}{7}$ and $\dfrac{x-2}{1} = \dfrac{y-b}{4} = \dfrac{z-2a}{7}$ lies on $xy$-plane, then the value of $a + b$ is :
Let $\vec{a}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ and $\vec{c}=\vec{a} \times \vec{b}$. Let $\vec{d}$ be a vector such that $|\vec{d}-\vec{a}|=\sqrt{11},|\vec{c} \times \vec{d}|=3$ and the angle between $\vec{c}$ and $\vec{d}$ is $\frac{\pi}{4}$. Then $\vec{a} \cdot \vec{d}$ is equal to
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