Download App

Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry1 Mark
MCQ
(2, -3, -1) 2x - 3y + 6z + 7 = 0:
  • A
    4
  • B
    3
  • C
    2
  • D
    $\frac{1}{5}$
✓

Answer

  1. 2

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Three Dimensional Geometry→Three Dimensional Geometry — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $f(x)=x^3+a x^2+b x+c$ where $a, b, c$ are real numbers. If $f(x)$ has a local minimum at $x=1$ and a local maximum at $x=-\frac{1}{3}$ and $f(2)=0$, then $\int_{-1}^1 f(x) d x$ equals
→
If in a $\triangle\text{ABC}$, $\text{A}=(0,0),\ \text{B}=(3,3\sqrt3),\ \text{C}=(-3\sqrt3,3)$, then the vecctor of magnitude $2\sqrt2$ units directed along AO, where O is the circumcenter of $\triangle\text{ABC}$ is,
  1. $(1-\sqrt3)\hat{\text{i}}+(1+\sqrt3)\hat{\text{j}}$
  2. $(1+\sqrt3)\hat{\text{i}}+(1-\sqrt3)\hat{\text{j}}$
  3. ​​​​​​​​​​​​​​$(1+\sqrt3)\hat{\text{i}}+(\sqrt3-1)\hat{\text{j}}$
  4. None of these
→
${d \over {dx}}\left( {{x^3}{{\tan }^2}{x \over 2}} \right)=$
→
Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}|=1$ and $|\vec{b} \times \vec{a}|=2$. Then $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is equal to
→
Function $f\left( x \right) = \left( {x - 2} \right)\left| {x - 3} \right|$ is monotonically increasing in
 
→
Fifteen coupons are numbered 1 to 15. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9 is:
→
Consider the function

$f(x)=\frac{\mathrm{P}(\mathrm{x})}{\sin (\mathrm{x}-2)}, \quad \mathrm{x} \neq 2$

$\quad \quad \quad \quad 7, \quad\quad\quad \mathrm{x}=2$

where $P(x)$ is a polynomial such that $P^{\prime \prime}(x)$ is always a constant and $P(3)=9$. If $f(x)$ is continuous at $x=2$, then $P(5)$ is equal to $.....$

→
The region formed by the inequalities $2 x+3 y-5 \leq 0,4 x-3 y+2 \leq 0$ and $x \geq 0 \ldots \ldots \ldots$
→
The solution of the differential equation $(x - {y^2}x)dx = (y - {x^2}y)dy$ is
→
If $f(x) = \int_{ - 1}^x {|t|\,dt,} $ $x \ge - 1,$ then
→