Introduction to Three Dimensional Geometry Test - 42

Let image of the point \(P(1,3,4)\) is \(Q\) in the given plane.

The equation of the line through \(P\) and normal to the given plane is 

\(\frac{(x-1) }{ 2}=\frac{(y-3) }{-1}=\frac{(z-4) }{ 1}=r\) (say)

Since the line passes through Q. 

So, let the coordinate of \(Q\) are \((2 r+1,-r\) \(+3, r+4)\).

Now, the coordinate of the mid-point of \(P Q\) is \((r+1,-\frac{r }{ 2}+3, \frac{r }{ 2}+4)\).

Now, this point lies in the given plane.

\(2(r+1)-(-\frac{r }{ 2}+3)+(\frac{r }{ 2}+4)+3=0 \)

\(\Rightarrow 2 r+2+\frac{r}{ 2}-3+\frac{r }{ 2}+4+3=0 \)

\(\Rightarrow 3 r+6=0 \)

\(\Rightarrow r=-2\)

So, the coordinate of \(Q\) is \((2 r+1,-r+3, r+4)=(-4+1,2+3,-2+4)\) \(=(-3,5,2)\).

Let the line segment be represented in vector form as \(\overrightarrow{ a }= a _{1} \overrightarrow{ i }+ a _{2} \overrightarrow{ j }+ a _{3} \overrightarrow{ k }\).

Vector along coordinate axes are \(\vec{ i }, \vec{ j }, \vec{ k }\).

Given that projection of \(\vec{a}\) on \(x\) axis is 12, that with y axis is 4 and that with \(z\) axis is 3.

\(\Rightarrow \vec{ a } \vec{ i }=12\)

\( \therefore a _{1}=12\)

Similarily, \(a _{2}=4, a _{3}=3\)

\(\therefore \overrightarrow{ a }=12 \overrightarrow{ i }+4 \overrightarrow{ j }+3 \overrightarrow{ k }\)

The length of the line segment \(=|\vec{a}|=\sqrt{144+16+9}=13\)

\(\therefore| P Q |=13\)

Thus the direction cosines of \(PQ\) are \(\frac{12}{13}, \frac{4}{13}, \frac{3}{13}\).

The lines \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}\) are mutually perpendicular.

As we know, if \(a, b, c\) are the direction ration ratios of a line passing through the point \(\left(x_{1}, y_{1}, z_{1}\right)\), then the equation of line is given by,

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

Let, \(L_{1}\) be \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\) and \(L_{2}\) be \(=\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}\) respectively.

Now by comparing \(L_{1}\) and \(L_{2}\) with \(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\), we get

\( a _{1}=-3, b _{1}=2 k , c _{1}=2, a _{2}=3 k , b _{2}=1\) and \(c _{2}=-5\)

As we know that if two lines are perpendicular, then

\(a _{1} \cdot a _{2}+ b _{1} \cdot b _{2}+ c _{1} \cdot c _{2}=0\)

\(\therefore (-3) \cdot 3k+2k \cdot 1+2 \cdot (-5)=0\)

\(\Rightarrow -9 k+2 k-10=0 \)

\(\Rightarrow-7 k=10\Rightarrow k=-\frac{10 }{ 7}\)

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 \(A (2,1,5)\), \(B (3,2,3)\) and \(C (4,0,4)\).

Now using the distance formula, we have

Distance between \(AB=\sqrt{(3-2)^{2}+(2-1)^{2}+(3-5)^{2}}\)

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

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

\(=\sqrt{6}\)

Distance between \(BC=\sqrt{(4-3)^{2}+(0-2)^{2}+(4-3)^{2}}\)

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

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

\(=\sqrt{6}\)

Distance between \(AC=\sqrt{(4-2)^{2}+(0-1)^{2}+(4-5)^{2}}\)

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

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

\(=\sqrt{6}\)

As, \(AB =BC =AC =\sqrt{6}\)

Therefore, it is a equilateral triangle and as we know that the circumcentre and orthocentre in equilateral triangle coincide each other.

So, the distance between them is \(0\).

As we know,

Equation of a plane parallel to a given plane ax + by + cz + d = 0 is ax + by + cz + λ = 0, where λ is a constant.

Equation of a plane parallel to a given plane 3x + 4y - 5z = 0 is 3x + 4y - 5z + λ = 0.

The plane is passing through the point (1, 2, 3).

So, point (1, 2, 3) satisfy the equation of plane 3x + 4y - 5z + λ = 0.

∴(3 × 1) + (4 × 2) - (5 × 3) + λ = 0

⇒ 3 + 8 - 15 + λ = 0

⇒ -4 + λ = 0

∴ λ = 4

So, Equation of plane is 3x + 4y - 5z + 4 = 0.

Let \(P (x ,y ,z)\) be the required point. Then

\(OP= AP=BP=CP\)

As we know,

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

\(\therefore\) The distance between O \((0,0,0)\) and \(P (x ,y ,z)\) is given by,

\(OP=\sqrt{(x - 0)^{2}+(y-0)^{2}+(z-0 )^{2}}=\sqrt{ x ^{2}+ y ^{2}+ z ^{2}}\)

The distance between \(A(a,0,0)\) and \(P (x ,y ,z)\) is given by,

\(AP=\sqrt{(x- a )^{2}+(y-0 )^{2}+(z-0)^{2}}=\sqrt{(x-a )^{2}+ y ^{2}+ z ^{2}}\)

The distance between \(B(0,b,0)\) and \(P (x ,y ,z)\) is given by,

\(BP=\sqrt{(x- 0)^{2}+(y-b )^{2}+(z-0)^{2}} =\sqrt{ x ^{2}+( y-b )^{2}+ z ^{2}}\)

The distance between \(C(0,0,c)\) and \(P (x ,y ,z)\) is given by,

\(CP=\sqrt{(x- 0)^{2}+(y-0 )^{2}+(z-c)^{2}} =\sqrt{ x ^{2}+ y ^{2}+(z-c )^{2}}\)

As, \(OP= AP\)

\(\therefore\sqrt{ x ^{2}+ y ^{2}+ z ^{2}}=\sqrt{( x-a)^{2}+ y ^{2}+ z ^{2}}\)

\(\Rightarrow x ^{2}+ y ^{2}+ z ^{2}=(x-a )^{2}+ y ^{2}+ z ^{2}\)

\(\Rightarrow x ^{2}=( a - x )^{2}\)

\(\Rightarrow x = a - x\)

\(\Rightarrow x+x= a\)

\(\Rightarrow 2x= a\)

\(\Rightarrow x =\frac{ a }{2}\)

Similarly, \(OP=BP\) and \(OP=CP\)

\(\therefore y = \frac{b}{2}\) and \(z =\frac{c}{2}\)

So, the required point has the coordinates \(\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2} \right)\).

Let the ratio be \(\lambda: 1\).

The dividing plane is \(YZ\) plane.

So, the co-ordinate of \(x\) after dividing the line \(=0\)

According to internal division rule of line segment,

\(x=\frac{3 \times \lambda-1 \times 2}{\lambda+1} \)

\(\Rightarrow \frac{3 \lambda-2}{\lambda+1}=0 \)

\(\Rightarrow 3 \lambda-2=0\)

\( \Rightarrow \lambda=\frac{2}{3}\)

So, Ratio \(=\lambda:1=\frac{2}{3}:1=2: 3\)

Given,

Equation of line: \(\frac{x-1}{3}=\frac{y+1}{-1}=\frac{z-3}{2}\)

Equation of plane: \(3 x+4 y+z+5=0\)

As we know,

The angle between the line \(\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}\) and the plane \(a_{2} x+b_{2} y+c_{2} z+d=0\) is given by,

\(\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\left(\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\right)\left(\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}\right)}\)

On comparing given equation of line and plane with above equation, we get

\(a_{1}=3, b_{1}=-1, c_{1}=2, a_{2}=3, b_{2}=4\) and \(c_{2}=1\)

\(\sin \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\left(\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\right)\left(\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}\right)}\)

\(\Rightarrow \sin \theta=\frac{3\times 3 +(-1)\times 4+2 \times 1}{\left(\sqrt{(3)^{2}+(-1)^{2}+(2)^{2}}\right)\left(\sqrt{(3)^{2}+(4)^{2}+(1)^{2}}\right)}\)

\(\Rightarrow \sin \theta=\frac{9 -4+2 }{\left(\sqrt{9+1+4}\right)\left(\sqrt{9+16+1}\right)}\)

\(\Rightarrow \sin \theta=\frac{7}{\sqrt{14} \cdot \sqrt{26}}\)

\(\Rightarrow \sin \theta=\sqrt{\frac{7}{52}} \)

\(\Rightarrow \theta=\sin ^{-1}\left(\sqrt{\frac{7}{52}}\right)\)

Given,

\(A (1,2,-1), B (2,5,-2), C (4,4,-3)\), and \(D (3,1,-2)\)

\(AB =\sqrt{(2-1)^{2}+(5-2)^{2}+(-2+1)^{2}}=\sqrt{11}\)

\(BC =\sqrt{(4-2)^{2}+(4-5)^{2}+(-3+2)^{2}}=\sqrt{6}\)

\(CD =\sqrt{(3-4)^{2}+(1-4)^{2}+(-2+3)^{2}}=\sqrt{11}\)

\(DA =\sqrt{(1-3)^{2}+(2-1)^{2}+(-1+2)^{2}}=\sqrt{6}\)

\(AB = CD\) and \(BC = DA\)

i.e., opposite sides are equal.

Now Direction ratio of \(AB =(1,3,-1)\) and Direction ratio of \(BC =(2,-1,-1)\)

\(\overrightarrow{ AB }=\hat{ i }+\hat{ j }+\hat{ k }\) and \(\overrightarrow{ BC }=2 \hat{ i }-\hat{ j }-\hat{ k }\)

\(\overrightarrow{ AB } \cdot \overrightarrow{ BC }=2-1-1=0\)

As the dot product between \(AB\) and \(BC\) is \(0\), that means angle between them is \(90^{\circ}\).

Similarly,on checking\(BC\) and \(CD\), we find that all angles are equal to \(90^{\circ}\) and opposite sides are equal.

So, it is a rectangle.