MCQ
A plane passing through (−1, 2, 3) and whose normal makes equal angle with the coordinate axes is:
  • A
    x + y + z + 4 = 0
  • B
    x − y + z + 4 = 0
  • x + y + z − 4 = 0
  • D
    x + y + z = 0

Answer

Correct option: C.
x + y + z − 4 = 0
Since normal makes equal angles with coordinate axis.

So, it intercept with all the axis will be same. So equation of plane will be

$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}+\frac{\text{x}}{\text{a}}=1$

⇒ x + y + z = a

Now, it passes through (−1, 2, 3), so

−1 + 2 + 3 = a

⇒ a = 4

⇒ x + y + z − 4 = 0

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

Similar questions

If $ \cos^{-1}\left (\frac {1 - \text{x}^{2}}{1 + \text{x}^{2}}\right ) + \cos^{-1}\left (\frac {1 - \text{y}^{2}}{1 + \text{y}^{2}}\right )=\frac{\pi}{2}$, where xy < 1, then:
If three conterminous edges of a parallelopiped are represented by $a - b,\,\,b - c$ and $c - a$, then its volume is
Choose the correct answer from the given four option.
The differential equation for $\text{y}=\text{A}\cos\alpha\text{x}+\text{B}\sin\alpha\text{x},$ where A and B are arbitrary constants is:
If $\text{y}=\frac{2}{\sqrt{\text{a}^2-\text{b}^2}}\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\tan\frac{\text{x}}{2}\Big),\text{a} > \text{b} > 0,$ then :
$\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big\{\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{a}}-\vec{\text{b}}-\vec{\text{c}}\big)\big\}$ is equal to:
Let $\text{f(x)=}\begin{cases}\frac{\text{x}-4}{|\text{x}-4|}+\text{a},&\text{if }\text{ x} < 4\\\text{a}+\text{b},&\text{if }\text{ x} =4\\\frac{\text{x}-4}{|\text{x}-4|}+\text{b},&\text{if }\text{ x} > 4\end{cases}$ Then, f(x) is continus at x = 4 when:
If a relation $R$ on the set $\{1,2,3\}$ be defined by $R=\{(1,2)\}$, then $R$ is
If matrix $A$ and $B$ has order respectively $m \times n$ and $n \times p$ then order of AB is :
If $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}$, where $\alpha, \beta$ and $\gamma$ are rational numbers, then $3 \alpha+4 \beta-\gamma$ is equal to ..........

 

Consider the matrices $A =$ $\left[ {\begin{array}{*{20}{c}}4&6&{ - 1}\\3&0&2\\1&{ - 2}&5\end{array}} \right]$ , $B =$ $\left[ {\begin{array}{*{20}{c}}2&4\\0&1\\{ - 1}&2\end{array}} \right]$ , $C =$ $\left[{\begin{array}{*{20}{c}}3\\1\\2\end{array}} \right]$ . Out of the given matrix products
$(i)$ $(AB)^TC$           $(ii)$ $C^TC(AB)^T$          $(iii)$ $C^TAB$        and       $(iv)$ $A^TABB^TC$