Question

Draw the projections of a point lying 20 mm above HP and is in first quadrant; if its shortest distance from the line of intersection of HP and VP is 40 mm. Also find the distance of the point from VP​

Answer 1

Answer:

Refer to the attachment please !!

Answer 2

Answer:

The distance of the point from VP​ is 20 mm.

Explanation:

Step 1: x and y are perpendicular vectors, so there will be no projection of x on y.

A point p which is AB=20 mm below the horizontal plane.

the shortest distance from xy i.e. origin=OB=40 mm

forms a right-angled triangle.

O B^2=O A^2+A B^2 (By Pythagoras theorem) ).

40^2=O A^2+20^2

1600-400=O A^2

O A^2=1200

$O A=\sqrt{1200}=\sqrt{10 * 10 * 2 * 2 * 3}$

$=10 * 2 * \sqrt{3}=20 \sqrt{3} \mathrm{~mm}$

the vector is $x=20 \sqrt{3} i$ and $y=-20 j_{-}$

Step 2:The two vectors $\mathrm{x}$ and $\mathrm{y}$ are perpendiculars

Projections of AB on $\mathbf{O A}=\frac{x, y}{\text { magnitudeof } x}=0$

As $I_1 \mathrm{j}=0$, therefore there is no projection of AB on OA.

But if you consider the whole vector $O B=O A+A B=x i-y j=20 \sqrt{3} i-20 j$

horizontal component $=40 \cos 30^0=40 * \sqrt{3} / 2=20 \sqrt{3}$

and vertical component is $=40 \sin 30^{\circ}=40 * 1 / 2=20$ along negative y axis.

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