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Chapter 11 Three Dimensional Geometry

In the previous class, we studied the introduction of 3d geometry. Now, we will go in depth of its concepts. This chapter consists of direction cosines and direction ratios of a line joining two points, coplanar and skew lines cartesian and vector equation of a line, shortest distance between two lines, cartesian and vector equation of a plane, angle between two lines; two planes; a line and a plane, distance of a point from a plane.

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Exercise 1

  • Q1

    If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines.

     

    Ans:

    Let direction cosines of the line be l, m, and n.

    \begin{align}l = cos90^0=0\end{align}

     \begin{align}m = cos135^0=-\frac{1}{\sqrt{2}}\end{align}

     \begin{align}n = cos45^0=\frac{1}{\sqrt{2}}\end{align}

    \begin{align}Therefore, the\; direction\; cosines\; of \;the\; line\; are\;0, -\frac{1}{\sqrt{2}}\;and\;\frac{1}{\sqrt{2}}\end{align}


    Q2

    Find the direction cosines of a line which makes equal angles with the coordinate axes.

    Ans:

    Let the direction cosines of the line make an angle α with each of the coordinate axes.

    l = cos α, m = cos α, n = cos α

         l2+m2+n2 =1

    ⇒ cos2α + cos2α + cos2α = 1

    ⇒ 3cos2α =1

    \begin{align}\Rightarrow cos^2α = \frac{1}{3}\end{align}

    \begin{align}\Rightarrow cosα = \pm\frac{1}{\sqrt 3}\end{align}

    Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are 

    \begin{align} \pm\frac{1}{\sqrt 3},\pm\frac{1}{\sqrt 3},and \pm\frac{1}{\sqrt 3}\end{align}


    Q3

    If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

    Ans:

    If a line has direction ratios of −18, 12, and −4, then its direction cosines are

    \begin{align} \frac{-18}{\sqrt {(-18)^2 + (12)^2 + (-4)^2}},\frac{12}{\sqrt {(-18)^2 + (12)^2 + (-4)^2}},\frac{-4}{\sqrt {(-18)^2 + (12)^2 + (-4)^2}}\end{align}

    \begin{align} i.e., \frac{-18}{22},\frac{12}{22},\frac{-4}{22}\end{align}

    \begin{align}  \frac{-9}{11},\frac{6}{11},\frac{-2}{11}\end{align}

    Thus, the direction cosines are 

    \begin{align} \frac{-9}{11},\frac{6}{11} and \frac{-2}{11}\end{align}


    Q4

    Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

    Ans:

    The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).

    It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2), are given by, x2x1, y2y1, and z2z1.

    The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3.

    The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6.

    It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional.

    Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B, and C are collinear.


    Q5

    The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

    Ans:

    The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

    vertices of triangle ABC

    The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.

    direction cosines of AB

    Therefore, the direction cosines of AB are

    direction of cosines of AB

    The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4.

    Therefore, the direction cosines of BC are

    direction cosines of BC

    The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., −8, −10, and 2.

    Therefore, the direction cosines of AC are

    direction cosines of AC


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