Download App

The plane — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsThe plane5 Marks
Question
State when the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ is parallel to the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}.$ Show that the line $\vec{\text{r}}=\hat{\text{i}}+\hat{\text{j}}+\lambda(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})$ is parallel to the plane $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3.$ Also, find the distance between the line and the plane.
✓

Answer

We know that line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$ is paralle to plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ if
$\vec{\text{b}}\cdot\vec{\text{n}}=0$
Given, line is $\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}})+\lambda(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})$ and plane is $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3,$
$\vec{\text{b}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\vec{\text{a}}=(\hat{\text{i}}+\hat{\text{j}})$ and $\vec{\text{n}}=(2\hat{\text{j}}+\hat{\text{k}})$
Now, $\vec{\text{b}}\cdot\vec{\text{n}}=(3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})(2\hat{\text{j}}+\hat{\text{k}})$
$=(3)(0)+(-1)(2)+(2)(1)$
$=0-2+2$
$=0$
Since, $\vec{\text{b}}\cdot\vec{\text{n}}=0$ So line is parallel to plane.
Distance between point $\vec{\text{a}}$ and plane $​​\vec{\text{r}}\cdot\vec{\text{n}}-\text{d}=0$ is given by
$\text{D}=\Bigg|\frac{\vec{\text{a}}\vec{\text{n}}-\text{d}}{|\vec{\text{n}}|}\Bigg|\ ...(\text{i})$
$\vec{\text{a}}$ is a point on the line. So diatance between line and plane is equal to the distance between $​​\vec{\text{a}}=(\hat{\text{i}}+\hat{\text{j}})$ and plane $\vec{\text{r}}\cdot(2\hat{\text{i}}+\hat{\text{k}})=3,$ so using (i)
$\text{D}=\Bigg|\frac{(\hat{\text{i}}+\hat{\text{j}})(2\hat{\text{j}}+\hat{\text{k}})-3}{\sqrt{(2)^2+(1)^2}}\Bigg|$
$=\bigg|\frac{(1)(0)+(1)(2)+(0)(1)-3}{\sqrt{4+1}}\bigg|$
$=\Big|\frac{0+2+0-3}{\sqrt{5}}\Big|$
$=\Big|\frac{-1}{\sqrt{5}}\Big|$
$=\frac{1}{\sqrt{5}}\text{ unit}$
So, required distance $=\frac{1}{\sqrt{5}}\text{ unit}$

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 "Solve the Following Question.(5 Marks)" from The plane→The plane — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the point having the following position vectors is collinear:$\vec{\text{a}},\ \vec{\text{b}},\ 3\vec{\text{a}}-2\vec{\text{b}}$
→
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\cot\text{x}}\text{ dx}$
→
Differentiate the following functions with respect to x:
$\log_\text{x}3$
→
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$\text{f}(\text{x})=\sqrt{25-\text{x}^2}\text{ on }[-3,4]$
→
If $\text{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix},$ show that $\text{A}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}$ and $\text{A}^3=\begin{bmatrix}1&3\\0&1\end{bmatrix}.$
→
A school wants to award its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs. 6,000. Three times the award money for Hard work added to that given for honesty amounts to Rs. 11,000. The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.
→

Image
Let u(x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v'(1) and w'(1). if it doesn’t exist then explain why?
→
$\int(\text{x}+2)\sqrt{3\text{x}+5}\text{ dx}$
→
A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by
$\text{M}=\frac{\text{WL}}{2}\text{x}-\frac{\text{W}}{2}\text{x}^{2}$
Find the point at which M is maximum in each case.
→
If $\text{y}=\log\sqrt{\frac{1+\tan\text{x}}{1-\tan\text{x}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\sec2\text{x}$
→