Given,
r×b+b×c=0
⇒r×b−c×b=0
⇒(r−c)×b=0
⇒(r−c)∣∣b
Therefore, r−c=λb
⇒r=c+λb
Also,
r⋅a=0 (given)
⇒(c+λb)⋅a=0
⇒c⋅a+λb⋅a=0
⇒λ=b⋅a−c⋅a
Now,
r⋅c=(c+λb)⋅c
=(c−b⋅ac⋅ab)⋅c
=∣c∣2−(b⋅ac⋅a)(b⋅c)
=74−(315)×8
=74−40=34
JEE Main 2023 — Mathematics Vectors & 3D Geometry
If a=i^+2k^,b=i^+j^+k^,c=7i^−3j^+4k^,r×b+b×c=0 and r⋅a=0 then r.c is equal to:
Held on 29 Jan 2023 · Verified 6 Jul 2026.
34
12
36
30
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
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
Work through every JEE Main Vectors & 3D Geometry PYQ, year by year.