Introduction to Three Dimensional Geometry Test - 41

Given lines are (on factorising)

lx+my=0, lx+my+n=0

lx−my=0, lx+my−n=0

Area \(=\left|\frac{\left(c_{1}-d_{1}\right)\left(c_{2}-d_{2}\right)}{\left(a_{1} b_{2}-a_{2} b_{1}\right)}\right|=\left|\frac{(0-n)(0+n)}{(-l m-l m)}\right|=\frac{n^{2}}{2|l m|}\)

As we know,

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

Given,

Two points are \((3 \sin \theta, 0,0)\) and \((4 \cos \theta, 0,0)\).

Distance between them \(=\sqrt{(4 \cos \theta-3 \sin \theta)^{2}+(0-0)^{2}+(0-0)^{2}}\)

\(=\sqrt{(4 \cos \theta-3 \sin \theta)^{2}}\)

\(=4 \cos \theta-3 \sin \theta\)...(1)

Now, maximum value of \(4 \cos \theta-3 \sin \theta=\sqrt{(4)^{2}+(-3)^{2}}\)

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

\(=\sqrt{25}\)

\(=5\)

From equation (1), we get

Distance between them \(=5\)

So, the maximum distance between points \((3 \sin \theta, 0,0)\) and \((4 \cos \theta, 0,0)\) is \(5\).

As we know,

The equation of the line that passes through the two points \(P\left(x_{1}, y_{1}, z_{1}\right)\) and \(Q\left(x_{2}, y_{2}, z_{2}\right)\) is given by the formula,

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

The line passing through the points \((1,2,-1)\) and \((3,-1,2)\) is,

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

Let, \(\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{3}=\lambda \)

\(\therefore x=2 \lambda+1, y=-3 \lambda+2 \text { and } z=3 \lambda-1\)

Line meets the yz-plane,

So, \(x =0 \)

\(\Rightarrow 2 \lambda+1=0\)

\(\Rightarrow 2 \lambda=-1\)

\(\Rightarrow \lambda=\frac{-1 }{ 2}\)

Now,

\(y=-3 \lambda+2\)

\(\Rightarrow y=-3 \times \frac{-1}{2}+2\)

\(\Rightarrow y =\frac{3}{2}+2\)

\(\Rightarrow y=\frac{7 }{ 2}\)

Now,

\(z=3 \lambda-1\)

\(\Rightarrow z=3 \times \frac{-1}{2}-1\)

\(\Rightarrow z =\frac{-3}{2}-1\)

\(\Rightarrow z=\frac{-5 }{ 2}\)

\(\therefore\) Point \((x, y, z)\) is \(\left(0, \frac{7}{2},-\frac{5}{2}\right)\).

Equation of the line through \((1,-2,3)\) parallel to the line \(\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}\) is:

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

Then any point on (1) is \((2 r +1,3 r -2,-6 r +3)\).

If this point lies on the plane \(x-y+z=5\), then

\((2 r+1)-(3 r-2)+(-6 r+3)=5\)

\(\Rightarrow 2r+1-3r+2-6r+3=5\)

\(\Rightarrow-7 r +6=5\)

\(\Rightarrow 6-5=7r\)

\(\Rightarrow 1=7r\)

\(\therefore r =\frac{1}{7}\)

So, the point \(=(2\times \frac{1}{7} +1,3\times \frac{1}{7}-2,-6 \times \frac{1}{7} +3)\)

\(=(\frac{2}{7} +1,\frac{3}{7}-2,\frac{-6}{7} +3)\)

\(=(\frac{2+7}{7},\frac{3-14}{7},\frac{-6+21}{7}\)

\(=\left(\frac{9}{7}, \frac{-11}{7}, \frac{15}{7}\right)\).

Distance between \((1,-2,3)\) and \(\left(\frac{9}{7}, \frac{-11}{7}, \frac{15}{7}\right)=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}\)

\(=\sqrt{\frac{49}{49}}\)

\(=\sqrt{1}\)

\(=1\)

Given,

The point \(P\) is on the \(y\)-axis.

Let the point P be \((0, y , 0)\), \(A\) be \((1,2,3)\) and \(B\) be \((2,3,4)\).

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 \(A(1,2,3)\) and \(P (0 ,y ,0)\) is given by,

\(AP=\sqrt{(0-1)^{2}+(y-2)^{2}+(0-3)^{2}}\)

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

\(=\sqrt{1+ (y-2)^{2}+ 9}\)

The distance between \(B(2,3,4)\) and \(P (0 ,y ,0)\) is given by,

\(BP=\sqrt{(0-2)^{2}+(y-3)^{2}+(0-4)^{2}}\)

\(=\sqrt{(-2 )^{2}+ (y-3)^{2}+ (-4)^{2}}\)

\(=\sqrt{4+ (y-3)^{2}+16}\)

Since, \(P\) is equidistant from the \((1,2,3)\) and \((2,3,4)\).

\(\therefore AP=BP\)

\(\Rightarrow \sqrt{\left(1+( y -2)^{2}+9\right)}=\sqrt{\left(4+( y -3)^{2}+16\right)}\)

\(\Rightarrow \left(1+( y -2)^{2}+9\right)=\left(4+( y -3)^{2}+16\right)\)

\(\Rightarrow\left(10+ y ^{2}-4 y +4\right)=\left(20+ y ^{2}-6 y +9\right) \)

\(\Rightarrow y ^{2}-4 y +14= y ^{2}-6 y +29 \)

\(\Rightarrow -4y+6y=29-14\)

\(\Rightarrow 2 y =15 \)

\(\Rightarrow y =\frac{15}{2}\)

\(\therefore P_y=\frac{15}{2}\)

As we know,

Let two points be \(A\left(x_{1}, y_{1}, z_{1}\right)\) and \(B\left(x_{2}, y_{2}, z_{2}\right)\).

Distance between \(A\) and \(B =\sqrt{\left( x _{2}- x _{1}\right)^{2}+\left( y _{2}- y _{1}\right)^{2}+\left( z _{2}- z _{1}\right)^{2}}\)

Midpoint \(=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)\)

The perpendicular distance of the point \((1,2,3)\) from \(x\) axis is \(h\).

\( h =\) distance between \((1,2,3)\) and \((1,0,0)\)

\(\therefore h =\sqrt{(1-1)^{2}+(0-2)^{2}+(0-3)^{2}}\)

\(\Rightarrow h =\sqrt{(0)^{2}+(-2)^{2}+(-3)^{2}}\)

\(\Rightarrow h =\sqrt{0+4+9} \)

\(\Rightarrow h =\sqrt{13}\)

The perpendicular distance of the point \((1,2,3)\) from \(z\) axis is \(k\).

\( k =\) distance between \((1,2,3)\) and \((0,0,3)\)

\(\therefore k =\sqrt{(0-1)^{2}+(0-2)^{2}+(3-3)^{2}}\)

\(\Rightarrow k =\sqrt{(-1)^{2}+(-2)^{2}+(0)^{2}}\)

\(\Rightarrow k =\sqrt{1+4+0} \)

\(\Rightarrow k =\sqrt{5}\)

Now,

\(h k =\sqrt{13} \times \sqrt{5}=\sqrt{65}\)

Given,

Three planes are

x + y = 0...(i)

y + z = 0...(ii)

x + z = 0...(iii)

Adding these three planes, we get

2(x + y + z) = 0

⇒ x + y + z = 0...(iv)

Putting value of (x + y) in (iv), we get

0 + z = 0

⇒ z = 0

Putting value of (y + z) in (iv), we get

x + 0 = 0

⇒ x = 0

Putting value of (x + z) in (iv), we get

y + 0 = 0

⇒ y = 0

So, (x, y, z) = (0, 0, 0)

So, the three planes meet in a unique point.

As we know,

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

Let the points be \(A (1,0,0), B (-1,0,0)\) and \(P ( x , y , z )\).

Given,

\(PA + PB =10\)

\(\Rightarrow \sqrt{( x -1)^{2}+( y -0)^{2}+( z -0)^{2}}+\sqrt{( x +1)^{2}+( y -0)^{2}+( z -0)^{2}}=10 \)

\(\Rightarrow \sqrt{( x -1)^{2}+ y ^{2}+ z ^{2}}=10-\sqrt{( x +1)^{2}+ y ^{2}+ z ^{2}}\)

Squaring both sides, we get

\(( x -1)^{2}+ y ^{2}+ z ^{2}=100+( x +1)^{2}+ y ^{2}+ z ^{2}-20 \sqrt{( x +1)^{2}+ y ^{2}+ z ^{2}} \)

\(\Rightarrow-4 x -100=-20 \sqrt{( x +1)^{2}+ y ^{2}+ z ^{2}} \)

\(\Rightarrow x +25=5 \sqrt{( x +1)^{2}+ y ^{2}+ z ^{2}}\)

Again squaring both sides, we get

\(x^{2}+50 x+625=25\left(\left(x^{2}+2 x+1\right)+y^{2}+z^{2}\right) \)

\(\Rightarrow x^{2}+50 x+625=25x^{2}+50 x+25+25y^{2}+25z^{2} \)

\(\Rightarrow x^{2}-25x^{2}+50 x-50 x-25+625=25y^{2}+25z^{2} \)

\(\Rightarrow -24x^{2}+600=25y^{2}+25z^{2} \)

\(\therefore 24 x^{2}+25 y^{2}+25 z^{2}-600=0\) is the required locus.

As we know,

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

Let the point on \(y\)-axis is \(O (0, y , 0)\) and another point be \(A(2,3,-1)\).

Distance between \(OA=\sqrt{(2 -0)^{2}+(3-y)^{2}+(-1-0 )^{2}}\)

Given,

\(OA=3\)

Squaring both sides, we get

\(OA ^{2}=9\)

\(\Rightarrow \left(\sqrt{(2 -0)^{2}+(3-y)^{2}+(-1-0 )^{2}}\right)^2=9\)

\(\Rightarrow(2-0)^{2}+(3- y )^{2}+(-1-0)^{2}=9\)

\(\Rightarrow 4+(3- y )^{2}+1=9\)

\(\Rightarrow 5+(3- y )^{2}=9\)

\(\Rightarrow(3- y )^{2}=9-5\)

\(\Rightarrow y^{2}-6y+9=4\)

\(\Rightarrow y^{2}-6y+5=0\)

\(\Rightarrow (y-1)(y-5)=0\)

\(\Rightarrow y =1,5\)

So, the point is either \((0,1,0)\) or \((0,5,0)\).