Question
The unit vector perpendicular to the plane passing through points $\text{P}\big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big),\text{Q}\big(2\hat{\text{i}}-\hat{\text{k}}\big)$ and $\text{R}\big(2\hat{\text{j}}+\hat{\text{k}}\big)$ is:
  1. $2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
  2. $\sqrt{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
  3. $\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
  4. $\frac{1}{6}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$

Answer

  1. $\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
Solution:
The vector $\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}$ is perpendicular to the vectors $\overrightarrow{\text{PQ}}$  and $\overrightarrow{\text{PR}}.$
$\therefore$ Required unit vector $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$
Now,
$\overrightarrow{\text{PQ}}=\text{P.V}\text{ of }\text{Q}-\text{P.V}.\text{ of P}$
$=\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
$\overrightarrow{\text{PR}}=\text{P.V}\text{ of }\text{R}-\text{P.V}.\text{ of P}$
$=-\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
$\therefore\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&-3\\-1&3&-1 \end{vmatrix}$
$=8\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$
$=4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$\Rightarrow\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|=\sqrt{64+16+16}$
$=\sqrt{96}$
$=4\sqrt{6}$
Required unit vector $=\frac{\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}}{\big|\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}\big|}$
$=\frac{4\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)}{4\sqrt{6}}$
$=\frac{1}{\sqrt{6}}\big(2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$

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

The function $\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$ is of the following type:
  1. Even and increasing.
  2. Odd and increasing.
  3. Even and decreasing.
  4. Odd and decreasing.
The vector $(\cos\text{a}\cos\beta)\hat{\text{i}}+(\cos\text{a}\sin\beta)\hat{\text{j}}+(\sin\text{a})\hat{\text{k}}$is a:
  1. Null vector
  2. Unit vector
  3. Constant vector
  4. None of these
If the function $f(x)=\left\{\begin{array}{cc}\frac{\sin x^2}{x} ; & x \neq 0 \\ 0 ; & x=0\end{array}\right.$, is differentiable at $x=0$, then right hand derivative of $f(x)$ at $x=0$ is
The corner points of the feasible region determined by the following system of linear inequalities:
2x + y ≤ 10, x + 3y ≤ 15, x, y ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5).
Let Z = px + qy, where p.q > 0.
Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is:
The Image of the point (2, -1, 5) in the plane $\vec{\text{r}},\hat{\text{i}}=0$ is:
  1. (-2, -1, 5)
  2. (2, 1, -5)
  3. (-2, 1, -5)
  4. (2, 0, 0)
The value of $\frac{0.76\times0.76\times0.76+0.24\times0.24\times0.24}{0.76\times0.76-0.76\times0.24+ 0.24+0.24}$ is:
Choose the correct answer from the given four options.
Find the value of $\lambda$ such that the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ are orthogonal:
  1. $0$
  2. $1$
  3. $\frac{3}{2}$
  4. $-\frac{5}{2}$
$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{\text{dx}}{1+\cos2\text{x}}\text{dx}$ is equal to:
  1. 1
  2. 2
  3. 3
  4. 4
The set points where the function f(x) given by $\text{f(x)=}|\text{x}-3|\cos\text{x}$ is  diffrentiable, is:
  1. R
  2. R - {3}
  3. $(0,\infty)$
  4. None of these.
The most economical speed to run the train is