Introduction to Three Dimensional Geometry Test - 40

As we know,

The equation of a line with direction ratio \((a, b, c)\) that passes through the point \(\left(x_{1}, y_{1}, z_{1}\right)\) is given by the formula,

\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)

If two planes intersect each other, the intersection will always be a line.

Given,

\(x=3 z+4\) and \(y=2 z-3 \)

\(\Rightarrow x-4=3 z\) and \(y+3=2 z \)

\(\Rightarrow \frac{x-4}{3}=z\) and \(\frac{y+3}{2}=z \)

\(\Rightarrow \frac{x-4}{3}=\frac{y+3}{2}=\frac{z}{1}\)

If two planes intersect each other, the intersection will always be a line.

Now, direction ratios of the line are \(( 3,2,1)\).

Let the line joining the points \(P(1,2,3)\) and \(Q(-3,4,-5)\) is divided by the \(x y\)-plane at point \(R\) in the ratio \(k: 1\).

Now, the coordinate of \(R\) is \(x=\frac{(-3 k+1) }{(k+1)},y=\frac{(4 k+2) }{(k+1)},z=\frac{(-5 k+3) }{(k+1)}\)

Since \(R\) lies on the \(xy\)-plane. So, z-coordinate is zero.

\(\frac{(-5 k+3) }{(k+1)}=0\Rightarrow -5k+3=0 \)

\(\Rightarrow -5k=-3\Rightarrow k=\frac{3 }{ 5}\)

So, the ratio \(=\frac{3 }{ 5}: 1=3: 5\)

Given,

AD is internal bisector of \(\angle B A C\).

Ratio at \(D\) is \(c : b\) where

\(c = AB =\sqrt{(3-5)^{2}+(2-3)+(0-2)^{2}}\)

\(=\sqrt{4+1+4}\)

\(=\sqrt{9}\)

\(=3\)

\(b = AC =\sqrt{(3+9)^{2}+(2-6)^{2}+(0+3)^{2}}\)

\(=\sqrt{144+16+9}\)

\(=\sqrt{169}\)

\(=13\)

For point \(D\),

\(x =\frac{ c (-9)+ b (6)}{ c + b }=\frac{3(-9)+13(5)}{3+13}=\frac{38}{16}=\frac{19}{8}\)

\(y =\frac{ c (6)+ b (3)}{ c + b }=\frac{3(6)+13(3)}{3+13}=\frac{57}{16}\)

\(z =\frac{ c (-3)+ b (2)}{ c + b }=\frac{3(-3)+13(2)}{3+13}=\frac{17}{16}\)

So, point \(D\) is \(\left[\frac{19}{8}, \frac{57}{16}, \frac{17}{6}\right]\).

Given,

Two planes \(2 x-3 y+6 z-5=0\) and \(6 x-9 y+18 z+20=0\)

The plane \(6 x-9 y+18 z+20=0\) can be re-written as \(2 x-3 y+6 z+\frac{20 }{ 3}=0\).

So, the planes \(2 x-3 y+6 z-5=0\) and \(2 x-3 y+6 z+\frac{20 }{ 3}=0\) are parallel.

As we know that, distance between two parallel planes \(ax +b y+c z+d_{1}=0\) and \(a x+b y+c z+d_{2}=0\) is given by,

\(\left|\frac{d_{1}-d_{2}}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)

\(\therefore d_{1}=-5, d_{2}=\frac{20 }{ 3}, a=2, b=-3\) and \(c=6\)

So, the distance between the given parallel planes \(=\left|\frac{-5-\frac{20}{3}}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}\right|\)

\(=\left|\frac{\frac{-15-20}{3}}{\sqrt{4+9+36}}\right|\)

\(=\left|\frac{\frac{-35}{3}}{\sqrt{49}}\right|\)

\(=\left|\frac{-35}{7\times 3}\right|\)

\(=\frac{5}{3}\) units

Let the given points be \(A (1,2,3)\) and \(B (3,2,-1)\) and let the point equidistant from \(A\) and \(B\) be \(P ( x , y , z )\).

\(\therefore P A=P B\)

\(\Rightarrow \sqrt{( x -1)^{2}+( y -2)^{2}+( z -3)^{2}}=\sqrt{( x -3)^{2}+( y -2)^{2}+( z +1)^{2}}\)

Squaring both sides, we get

\(( x -1)^{2}+( y -2)^{2}+( z -3)^{2}=( x -3)^{2}+( y -2)^{2}+( z +1)^{2} \)

\( \Rightarrow x ^{2}+1-2 x + z ^{2}+9-6 z = x ^{2}+9-6 x + z ^{2}+1+2 z \)

\(\Rightarrow -2 x -6 z +10=-6 x +2 z +10 \)

\(\Rightarrow -2 x+6x-6z-2z+10-10=0\)

\(\Rightarrow 4 x -8 z =0 \)

\(\Rightarrow 4( x -2 z) =0 \)

\(\Rightarrow x -2 z =0\)

Let point on \(x\)-axis be \(P(3,0,0)\) and the other point be \(Q(3,12,5)\).

As we know,

Distance between two point \(=\sqrt{(x_2 - x_1)^{2}+(y_2- y_1)^{2}+(z_2- z_1 )^{2}}\)

The distance between the \(x\)-axis and point \((3,12,5)=\) Distance between \(PQ\)

\(=\sqrt{(3-3)^{2}+(12-0)^{2}+(5-0)^{2}} \)

\(=\sqrt{0^{2}+12^{2}+5^{2}}\)

\(=\sqrt{144+25} \)

\(=\sqrt{169}\)

\(=13\)

So, the distance between the \(x\)-axis and point \((3,12,5)\) is \(13\).

Given,

Equation is 3x – 4y = 0.

Here z = 0

So, the given equation 3x – 4y = 0 represents a plane containing Z-axis.

As we know,

Equation of line joining two points \(\left(x_{1}, y_{1}, z_{1}\right)\) and \(\left(x_{2}, y_{2}, z_{2}\right)\) is given as,

\(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)

A point lie on the line only when its coordinates satisfy the equation of line.

Given,

\(P=(5, y, z)\)

Equation of line joining \((3,2,-1)\) and \((6,-4,-2)\) is

\(\frac{x-3}{6-3}=\frac{y-2}{-4-2}=\frac{z+1}{-2+1}\)

\(\Rightarrow \frac{x-3}{3}=\frac{y-2}{-6}=\frac{z+1}{-1}\)

So, if point \(P\) lie on the line then it must satisfy the above equation.

\(\therefore \frac{5-3}{3}=\frac{y-2}{-6}=\frac{z+1}{-1} \)

\(\Rightarrow \frac{5-3}{3}=\frac{y-2}{-6}\)

\(\Rightarrow \frac{2}{3}=\frac{y-2}{-6}\)

\(\Rightarrow \frac{2\times-6}{3}=y-2\)

\(\Rightarrow -4+2=y\)

\(\therefore y=-2\)

So, y coordinate of \(P\) is \(-2\).

Let \(a\) is the length of the side of a square.

Given,

The diagonal of a square are \((1,-2,3)\) and \((2,-3,5)\).

Now, length of the diagonal of square \(=\sqrt{1-2)^{2}+(-2+3)^{2}+(3-5)^{2}}\)

\(=\sqrt{ 1+1+4}\)

\(=\sqrt{6}\)

As we know,

Length of diagonal of square \(=a\sqrt{2}\) (where \(a=\) side of square)

\(\therefore a\sqrt{2}=\sqrt{6} \)

\(\Rightarrow a \sqrt{2}=\sqrt{3} \times \sqrt{2} \)

\(\Rightarrow a=\sqrt{3}\)

So, the length of the side of the square is \(\sqrt{3}\) unit.