Answer
Verified
469.8k+ views
Hint: The projection of a point is its shadow on the plane or central projection.
If C is a point, called the centre of projection then the projection of a point P different from C onto a plane that does not contain C is the interaction of the line CP with the plane.
Complete step-by- step solution:
Let us draw a plane and the projection of point \[P(\mathop p\limits^ \to )\]on the plane is \[\mathop s\limits^ \to \].
The intersection is \[\mathop r\limits^ \to .\mathop n\limits^ \to = q\]
As the line is normal to the plane i.e. perpendicular to the plane and vector \[\mathop P\limits^ \to \] is passing through the plane and parallel to \[\mathop n\limits^ \to \]
\[E{q^n}\] of such a line is\[\overrightarrow r = \overrightarrow p + \lambda \overrightarrow n ……...(1)\]
Given, \[\mathop r\limits^ \to .\mathop n\limits^ \to = q........(2)\]
As the line is passing through the plane, then the equation (1) will be satisfying equation (2) and that point \[\mathop r\limits^ \to = \mathop s\limits^ \to \]
Substituting equation (1) in (2), we get:
$\Rightarrow$ \[(\mathop p\limits^ \to + \lambda \mathop n\limits^ \to )\mathop n\limits^ \to = q\]
To find the value of \[\lambda \], simplify the above term then we get it as
\[ \Rightarrow \mathop p\limits^ \to . \mathop n\limits^ \to + \lambda \mathop n\limits^ \to .\mathop n\limits^ \to = q\]
As \[\left[ {\overrightarrow n .\overrightarrow n = {{\left| {\mathop n\limits^ \to } \right|}^2}} \right]\] , we get:
\[ \Rightarrow \lambda {\left| {\mathop n\limits^ \to } \right|^2} = q - \mathop p\limits^ \to .\mathop n\limits^ \to \]
\[ \Rightarrow \lambda = \dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}}\]______ (3) {On RHS \[{\left| {\mathop n\limits^ \to } \right|^2}\] will be in division as it was multiplication on LHS}
Now using equation (3) in (1), we get:
\[\mathop r\limits^ \to = \mathop p\limits^ \to + (\dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}})\mathop n\limits^ \to \]
We know that\[\overrightarrow r = \overrightarrow s \], hence:
\[\mathop s\limits^ \to = \mathop p\limits^ \to + (\dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}})\mathop n\limits^ \to \]
Note: Two planes are parallel if they have the same normal vector (i.e. their normal vectors are parallel). If two planes are not parallel, then they intersect in a line.If any line passes through a plane then it always satisfies the equation of that plane.
If C is a point, called the centre of projection then the projection of a point P different from C onto a plane that does not contain C is the interaction of the line CP with the plane.
Complete step-by- step solution:
Let us draw a plane and the projection of point \[P(\mathop p\limits^ \to )\]on the plane is \[\mathop s\limits^ \to \].
The intersection is \[\mathop r\limits^ \to .\mathop n\limits^ \to = q\]
As the line is normal to the plane i.e. perpendicular to the plane and vector \[\mathop P\limits^ \to \] is passing through the plane and parallel to \[\mathop n\limits^ \to \]
\[E{q^n}\] of such a line is\[\overrightarrow r = \overrightarrow p + \lambda \overrightarrow n ……...(1)\]
Given, \[\mathop r\limits^ \to .\mathop n\limits^ \to = q........(2)\]
As the line is passing through the plane, then the equation (1) will be satisfying equation (2) and that point \[\mathop r\limits^ \to = \mathop s\limits^ \to \]
Substituting equation (1) in (2), we get:
$\Rightarrow$ \[(\mathop p\limits^ \to + \lambda \mathop n\limits^ \to )\mathop n\limits^ \to = q\]
To find the value of \[\lambda \], simplify the above term then we get it as
\[ \Rightarrow \mathop p\limits^ \to . \mathop n\limits^ \to + \lambda \mathop n\limits^ \to .\mathop n\limits^ \to = q\]
As \[\left[ {\overrightarrow n .\overrightarrow n = {{\left| {\mathop n\limits^ \to } \right|}^2}} \right]\] , we get:
\[ \Rightarrow \lambda {\left| {\mathop n\limits^ \to } \right|^2} = q - \mathop p\limits^ \to .\mathop n\limits^ \to \]
\[ \Rightarrow \lambda = \dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}}\]______ (3) {On RHS \[{\left| {\mathop n\limits^ \to } \right|^2}\] will be in division as it was multiplication on LHS}
Now using equation (3) in (1), we get:
\[\mathop r\limits^ \to = \mathop p\limits^ \to + (\dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}})\mathop n\limits^ \to \]
We know that\[\overrightarrow r = \overrightarrow s \], hence:
\[\mathop s\limits^ \to = \mathop p\limits^ \to + (\dfrac{{q - \mathop p\limits^ \to .\mathop n\limits^ \to }}{{{{\left| {\mathop n\limits^ \to } \right|}^2}}})\mathop n\limits^ \to \]
Note: Two planes are parallel if they have the same normal vector (i.e. their normal vectors are parallel). If two planes are not parallel, then they intersect in a line.If any line passes through a plane then it always satisfies the equation of that plane.
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
If the coordinates of the points A B and C be 443 23 class 10 maths JEE_Main
If the mean of the set of numbers x1x2xn is bar x then class 10 maths JEE_Main
What is the meaning of celestial class 10 social science CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE
Give 10 examples for herbs , shrubs , climbers , creepers