Three Dimensional Geometry Question Answers: NCERT Class 12 Mathematics

Exercise 1
Q:
A:

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}


Q:
A:

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}


Q:
A:

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}


Q:
A:

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.


Q:
A:

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