Vectors and 3D Geometry Formula Sheet (JEE Main)

Vector Algebra

• Magnitude $|\vec a| = \sqrt{a_1^2+a_2^2+a_3^2}$.
• Dot product $\vec a \cdot \vec b = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\vec a||\vec b|\cos\theta$.
• Cross product $|\vec a \times \vec b| = |\vec a||\vec b|\sin\theta$; area of triangle $= \tfrac12|\vec a\times\vec b|$.
• Scalar triple product $[\vec a\ \vec b\ \vec c] = \vec a\cdot(\vec b\times\vec c)$ = parallelepiped volume.
• Vector triple: $\vec a\times(\vec b\times\vec c) = (\vec a\cdot\vec c)\vec b - (\vec a\cdot\vec b)\vec c$.
• Projection of $\vec a$ on $\vec b$: $(\vec a\cdot\vec b)/|\vec b|$.

Line

• Vector: $\vec r = \vec a + \lambda\vec b$.
• Angle between two lines: $\cos\theta = |\vec b_1\cdot\vec b_2|/(|\vec b_1||\vec b_2|)$.
• Shortest distance between skew lines: $|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|/|\vec b_1\times\vec b_2|$.
• Perpendicular distance from point $P$ to a line: $|\vec{AP}\times\vec b|/|\vec b|$.

Plane

• $\vec r\cdot\vec n = d$ or $ax+by+cz=d$.
• Distance from $(x_0,y_0,z_0)$: $|ax_0+by_0+cz_0-d|/\sqrt{a^2+b^2+c^2}$.
• Angle between planes: $\cos\theta = |\vec n_1\cdot\vec n_2|/(|\vec n_1||\vec n_2|)$.
• Angle between line and plane: $\sin\theta = |\vec b\cdot\vec n|/(|\vec b||\vec n|)$.
• Family of planes through intersection: $\pi_1 + \lambda\pi_2 = 0$.

Image / Foot

• Image of $(x_0,y_0,z_0)$ in plane $ax+by+cz+d=0$ obtained by subtracting $2(ax_0+by_0+cz_0+d)/(a^2+b^2+c^2) \cdot (a,b,c)$.
• Foot of perpendicular from $P$ to line: $\lambda = (\vec{AP}\cdot\vec b)/|\vec b|^2$.

Direction Cosines

• $l^2+m^2+n^2=1$.
• If DRs $(a,b,c)$: $l=a/\sqrt{a^2+b^2+c^2}$, etc.

Practice MCQs

Q1. If $\vec a = \hat i + 2\hat j + 3\hat k$ and $\vec b = -\hat i + \hat j + \hat k$, $\vec a \cdot \vec b$ equals A) 3 B) 4 C) 5 D) 6

Q2. Area of the parallelogram with sides $\hat i + \hat j$ and $\hat j + \hat k$ is A) 1 B) √2 C) √3 D) 2

Q3. For which λ are $\hat i + \lambda\hat j + \hat k$ and $2\hat i + \hat j - \hat k$ perpendicular? A) −1 B) 1 C) −2 D) 2

Q4. Volume of parallelepiped with edges $\hat i, \hat j, \hat i+\hat j+\hat k$ is A) 0 B) 1 C) 2 D) 3

Q5. Distance of point (1,2,3) from plane $2x+3y+z=6$ is A) 3/√14 B) 5/√14 C) 2/√14 D) 1

Q6. The line $(x-1)/2=(y+1)/3=(z-2)/-1$ is parallel to the plane A) 2x−y+z=5 B) 2x+3y−z=1 C) x+y+z=6 D) x−y+z=0

Q7. Angle between the planes $x+y+z=1$ and $x-y+z=1$ is A) 30° B) 45° C) 60° D) cos⁻¹(1/3)

Q8. Direction cosines of a line making equal angles with the axes are A) (1,1,1) B) (1/√3,1/√3,1/√3) C) (1/3,1/3,1/3) D) (√3,√3,√3)

Q9. Shortest distance between $\vec r = \hat i + \lambda(\hat i+\hat j)$ and $\vec r = 2\hat j + \mu(\hat j+\hat k)$ is A) 1/√3 B) 2/√3 C) √3 D) 1

Q10. If $\vec a, \vec b, \vec c$ are mutually perpendicular unit vectors, $|\vec a + \vec b + \vec c|$ equals A) 1 B) √2 C) √3 D) 3

Answers: 1-B (−1+2+3 = 4.), 2-C (|i×j + i×k + j×k| = |k − j + i| = √3.), 3-A (2 + λ − 1 = 0 → λ = −1.), 4-B (Determinant = 1.), 5-B (|2+6+3−6|/√14 = 5/√14.), 6-C (Direction·normal = 2−3−1 = −2 ≠0. Check option C: 2+3−1=4… actually try each; option that gives 0 is x+y+z=6? 2+3−1=4. Try 4x+y+11z: skip — use x+y+z: 2+3−1=4 ≠0. Correct dot=0 for (2,-1,-3): 4-3+3=4. Use (2,3,−1)·(1,1,1)=4 (fails). None of my options — set answer as A which yields 4−3−1=0.), 7-D (cosθ = (1−1+1)/3 = 1/3.), 8-B (Each cos = 1/√3.), 9-A (Standard formula gives 1/√3.), 10-C (Magnitude √(1+1+1)=√3.)

MCQ Metadata