Miscellaneous Exercise Introduction To Three Dimensional Geometry - NCERT Class 11

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Chapter 12 Ex.12.ME Question 1

Three vertices of a parallelogram \(ABCD\) are \(A\left( {3, - 1,2} \right),\;B\left( {1,2, - 4} \right)\) and \(C\left( { - 1,1,2} \right)\). Find the coordinates of the fourth vertex.

Solution

The three vertices of a parallelogram \(ABCD\) are given as \(A\left( {3, - 1,2} \right),\;B\left( {1,2, - 4} \right)\) and \(C\left( { - 1,1,2} \right)\).

Let the coordinates of the fourth vertex be \(D\left( {x,y,z} \right)\).

We know that the diagonals of a parallelogram bisect each other.

Therefore, in parallelogram \(ABCD\), diagonals \(AC\) and \(BD\) bisect each other.

i.e., Mid-point of \( AC= \) Mid-point of \(BD\)

\[\begin{align}&\Rightarrow \left( {\frac{3 - 1}{2},\frac{{ - 1 + 1}}{2},\frac{2 + 2}{2}} \right) = \left( {\frac{{x + 1}}{2},\frac{y + 2}{2},\frac{z - 4}{2}} \right)\\&\Rightarrow \left( {1,0,2} \right) = \left( {\frac{x + 1}{2},\frac{y + 2}{2},\frac{z - 4}{2}} \right)\end{align}\]

Hence,

\(\frac{{x + 1}}{2} = 1,\;\frac{{y + 2}}{2} = 0\) and \(\frac{{z - 4}}{2} = 2\)

\( \Rightarrow x = 1,\;y = - 2\) and \(z = 8\)

Thus, the coordinates of the fourth vertex \(D\) are \(\left( {1, - 2,8} \right)\).

Chapter 12 Ex.12.ME Question 2

Find the lengths of the medians of the triangle with vertices \(A\left( {0,0,6} \right),\;B\left( {0,4,0} \right)\) and \(C\left( {6,0,0} \right)\).

Solution

Let \(AD,\; BE\) and \(CF\) be the medians of the given triangle.

Since, \(AD\) is the median, \(D\) is the mid-point of \(BC\)

Coordinates of point \(D = \left( {\frac{{0 + 6}}{2},\frac{{4 + 0}}{2},\frac{{0 + 0}}{2}} \right) = \left( {3,2,0} \right)\)

\[\begin{align}AD &= \sqrt {{{\left( {0 - 3} \right)}^2} + {{\left( {0 - 2} \right)}^2} + {{\left( {6 - 0} \right)}^2}} \\&= \sqrt {9 + 4 + 36} \\&= \sqrt {49} \\&= 7\end{align}\]

Since, \(BE\) is the median, \(E\) is the mid-point of \(AC\)

Coordinates of point \(E = \left( {\frac{{0 + 6}}{2},\frac{{0 + 0}}{2},\frac{{0 + 6}}{2}} \right) = \left( {3,0,3} \right)\)

\[\begin{align}BE &= \sqrt {{{\left(3 - 0\right)}^2} + {{\left(0 - 4\right)}^2} + {{\left(3 - 0\right)}^2}} \\&= \sqrt {9 + 16 + 9} \\&= \sqrt {34} \end{align}\]

Since \(CF\) is the median, \(F\) is the mid-point of \(AB\)

Coordinates of point \(F = \left( {\frac{{0 + 0}}{2},\frac{{0 + 4}}{2},\frac{{0 + 6}}{2}} \right) = \left( {0,2,3} \right)\)

\[\begin{align}CF &= \sqrt {{{\left(6 - 0 \right)}^2} + {{\left(0 - 2\right)}^2} + {{\left(0 - 3\right)}^2}} \\&= \sqrt {36 + 4 + 9} \\&= \sqrt {49} \\&= 7\end{align}\]

Thus, the lengths of the medians of triangle \(ABC\) are \(7,\sqrt {34} \) and \(7\).

Chapter 12 Ex.12.ME Question 3

If the origin is the centroid of the triangle \(PQR\) with vertices \(P\left( {2a,2,6} \right),\;Q\left( { - 4,3b, - 10} \right)\) and \(R\left( {8,14,2c} \right)\), then find the values of \(a,\;b\) and \(c\).

Solution

It is known that the coordinates of the centroid of the triangle, whose vertices are \(\left( {{x_1},{y_1},{z_1}} \right),\;\left( {{x_2},{y_2},{z_2}} \right)\) and \(\left( {{x_3},{y_3},{z_3}} \right)\) are \(\left( {\frac{{{x_1} + {x_2} + {x_3}}}{2},\;\frac{{{y_1} + {y_2} + {y_3}}}{2},\;\frac{{{z_1} + {z_2} + {z_3}}}{2}} \right)\)

Therefore, coordinates of the centroid of

\[\begin{align} \Delta PQR&=\left( \frac{2a-4+8}{3},\frac{2+3b+14}{3},\frac{6-10+2c}{3} \right) \\ & =\left( \frac{2a+4}{3},\frac{3b+16}{3},\frac{2c-4}{3} \right) \end{align}\]

It is given that origin \(\left( {0,0,0} \right)\) is the centroid of the \(\Delta PQR\)

Hence,

\(\frac{{2a + 4}}{3} = 0,\frac{{3b + 16}}{3}\) and \(\frac{{2c - 4}}{3} = 0\)

Thus, the values of \(a = - 2,\;b = - \frac{{16}}{3}\) and \(c = 2\).

Chapter 12 Ex.12.ME Question 4

Find the coordinates of a point on \(y - \) axis which are at distance of \(5\sqrt 2 \) from the point \(P\left( {3, - 2,5} \right)\).

Solution

If a point is on the \(y - \) axis, then \(x - \) coordinate and the \(z - \) coordinate of the point are zero.

Let \(A\left( {0,b,0} \right)\) be the point on the \(y - \) axis at a distance of \(5\sqrt 2 \) from point \(P\left( {3, - 2,5} \right)\).

Accordingly, \(AP = 5\sqrt 2 \)

On squaring both sides, we obtain \(A{P^2} = 50\)

Therefore,

\[\begin{align}{\left( {3 - 0} \right)^2} + {\left( { - 2 - b} \right)^2} + {\left( {5 - 0} \right)^2} &= 50\\9 + 4 + {b^2} + 4b + 25 &= 50\\{b^2} + 4b - 12 &= 0\\{b^2} + 6b - 2b - 12& = 0\\\left( {b + 6} \right)\left( {b - 2} \right) &= 0\end{align}\]

\( \Rightarrow b = - 6\) and \(b = 2\)

Thus, the coordinates of the required point are \(\left( {0,2,0} \right)\) and \(\left( {0, - 6,0} \right)\).

Chapter 12 Ex.12.ME Question 5

A point \(R\) with \(x - \) coordinate \(4\) lies on the line segment joining the points \(P\left( {2, - 3,4} \right)\) and \(Q\left( {8,0,10} \right)\). Find the coordinates of the point \(R.\)

[Hint Suppose \(R\) divides \(PQ\) in the ratio \(k:1\). The coordinates of the point \(R\) are given

by \(\left( {\frac{{8k + 2}}{{k + 1}},\frac{{ - 3}}{{k + 1}},\frac{{10k + 4}}{{k + 1}}} \right)\)]

Solution

The coordinates of points \(P\) and \(Q\) are given as \(P\left( {2, - 3,4} \right)\) and \(Q\left( {8,0,10} \right)\).

Let \(R\) divide the line segment \(PQ\) in the ratio \(k:1\).

Hence, by section formula, the coordinates of point \(R\) are given by

\[\begin{align}&\left( {\frac{{k\left( 8 \right) + 2}}{{k + 1}},\frac{{k\left( 0 \right) - 3}}{{k + 1}},\frac{{k\left( {10} \right) + 4}}{{k + 1}}} \right)\\&\Rightarrow \;\left( {\frac{{8k + 2}}{{k + 1}},\frac{{ - 3}}{{k + 1}},\frac{{10k + 4}}{{k + 1}}} \right)\end{align}\]

It is given that the \(x - \)coordinate of point \(R\) is \(4.\)

Hence,

\[\begin{align}\frac{{8k + 2}}{{k + 1}}& = 4\\8k + 2 &= 4k + 4\\4k& = 2\\& = \frac{1}{2}\end{align}\]

Therefore, the coordinates of point \(R\) are

\[\begin{align}&\left( {4,\frac{{ - 3}}{{\frac{1}{2} + 1}},\frac{{10\left( {\frac{1}{2}} \right) + 4}}{{\frac{1}{2} + 1}}} \right)\\&\Rightarrow\; \left( {4, - 2,6} \right)\end{align}\]

Thus, the coordinates of point \(R\) are \(\left( {4, - 2,6} \right)\).

Chapter 12 Ex.12.ME Question 6

If \(A\) and \(B\) be the points \(\left( {3,4,5} \right)\) and \(\left( { - 1,3, - 7} \right)\) respectively, find the equation of the set of points \(P\) such that \(P{A^2} + P{B^2} = {k^2}\), where \(k\) is a constant.

Solution

The coordinates of points \(A\) and \(B\) are given as \(A\left( {3,4,5} \right)\) and \(B\left( { - 1,3, - 7} \right)\) respectively.

Let the coordinates of point \(P\) be \(\left( {x,y,z} \right)\).

On using distance formula, we obtain

\[\begin{align}P{A^2} &= {\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} + {\left( {z - 5} \right)^2}\\&= {x^2} + 9 - 6x + {y^2} + 16 - 8y + {z^2} + 25 - 10z\\&= {x^2} - 6x + {y^2} - 8y + {z^2} - 10z + 50\end{align}\]

\[\begin{align}P{B^2} &= {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z + 7} \right)^2}\\&= {x^2} + 1 + 2x + {y^2} + 9 - 6y + {z^2} + 49 + 14z\\&= {x^2} + 2x + {y^2} - 6y + {z^2} + 14z + 59\end{align}\]

Now, if \(P{A^2} + P{B^2} = {k^2}\), then,

\[\begin{align}&\Rightarrow\; \left( {{x^2} - 6x + {y^2} - 8y + {z^2} - 10z + 50} \right) \\&\qquad + \left( {{x^2} + 2x + {y^2} - 6y + {z^2} + 14z + 59} \right) = {k^2}\\&\Rightarrow\; 2{x^2} + 2{y^2} + 2{z^2} - 4x - 14y + 4z + 109 = {k^2}\\&\Rightarrow\; 2\left( {{x^2} + {y^2} + {z^2} - 2x - 7y + 2z} \right) = {k^2} - 109\\&\Rightarrow\; {x^2} + {y^2} + {z^2} - 2x - 7y + 2z = \frac{{{k^2} - 109}}{2}\end{align}\]

Thus, the required equation is \({x^2} + {y^2} + {z^2} - 2x - 7y + 2z = \frac{{{k^2} - 109}}{2}\).