Practice AP Inter 2nd Year Maths Study Material Chapter 11 Three Dimensional Geometry MCQ to identify your strengths and weak areas.
AP Inter 2nd Year Maths Three Dimensional Geometry MCQ
Question 1.
If α, β, γ are the angles made by the line with positive direction of the coordinate axes, then sin2α + sin2β + sin2 γ =
1) 1
2) 2
3) 3
4) \(\frac{3}{2}\)
Solution:
2) 2
α, β, γ are angle made by the line with +ve direction of coordinate axes
1 = cosα, m = cosβ, n = cosγ are dc’s of the line ⇒ l2 + m2 + n = 1
⇒ cos2 α + cos2 β + cos2 γ = 1 ⇒ (1 – sin2 α) + (1 – sin2 β) + (1 – sin2 γ) = 1
⇒ 3 – 1 = sin2 α + sin2 β + sin2 γ ⇒ sin2 a + sin2 p + sin2 γ = 2
Question 2.
The direction cosines of the median of the triangle formed by A(1, -3, 2) B(3, 1, 2) and C(-1, 3, -3) which passing through the vertex C is
1) \(\left(\frac{3}{5 \sqrt{2}}, \frac{4}{5 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
2) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{1}{5 \sqrt{2}}\right)\)
3) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{\sqrt{2}}\right)\)
4) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right)\)
Solution:
3) \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{\sqrt{2}}\right)\)
Mid point of AB = F = \(\left(\frac{1+3}{2}, \frac{-3+1}{2}, \frac{2+2}{2}\right)\) = (2, -1, 2), C = (-1, 3, -3)
d.r’s of Median CF = (a, b, c) = (2 + 1, -1 – 3, 2 + 3) = (3, -4, 5) ⇒ \(\sqrt{9+16+25}=\sqrt{50}=5 \sqrt{2}\)
d.c’s = \(\left(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{5}{5 \sqrt{2}}\right)\)
Question 3.
If the line joining the points A(2, 3, 4) and B(3, -2, 2) is parallel to the line joining C(1, -2, z) and D(-1, y, -1), then y + z =
1) 13
2) 3
3) -3
4) -13
Solution:
2) 3
Given AB || CD ⇒ \(\left(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\right)=\frac{-2}{1}=\frac{y+2}{-5}=\frac{-1-z}{-2}\)
⇒ \(\frac{-2}{1}=\frac{y+2}{-5}=\frac{1+z}{2} \Rightarrow-2=\frac{y+2}{-5} \text { and }-2=\frac{1+z}{2}\) ⇒ -4 = 1 + z ⇒ -5 = z
⇒ 10 = y + 2 ⇒ 8 = y ⇒ y + z ⇒ 8 + (-5) = 3
Question 4.
If the two lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\lambda(\mathbf{P} \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \text { and } \overline{\mathbf{r}}=(4 \hat{\mathbf{i}}-\mathbf{P} \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(3 \mathbf{P} \hat{\mathbf{i}}+5 \mathrm{P} \hat{\mathbf{j}}+3 \hat{\mathbf{k}})\) are perpendicular then p =
1) 2
2) 3
3) 6
4) 2 or 3
Solution:
4) 2 or 3
Dr’s of line (1) are (p, -3, 6); Dr’s of line (2) are (3p, 5p, 3)
Given lines are perpendicular
⇒ a1a2 + b1b2 + c1c2 = 0 ⇒ 3p(p) + 5p(-3) + 18 = 0 ⇒ 3p2 – 15p + 18 = 0
⇒ p2 – 5p + 6 = 0 ⇒ (p – 2)(p – 3) = 0 ⇒ p = 2 (or) p = 3
Question 5.
It the two lines \(\frac{x+1}{2 k}=\frac{y-3}{3}=\frac{z-4}{-7}\) and \(\frac{x-1}{1}=\frac{y+1}{-3 k}=\frac{z+2}{2}\) are perpendicular, then k =
1) 2
2) 1
3) -2
4) 14/11
Solution:
3) -2
Given lines are perpendicular ⇒ 2k(1) + 3(-3k) + (-7)(2) = 0 ⇒ 2k – 9k – 14 = 0
⇒ -7k = 14 ⇒ k = -2
Question 6.
The angle between the lines \(\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+1}{1}\) and \(\frac{x+2}{1}=\frac{y+2}{1}=\frac{z-3}{2}\) is
1) \(\frac{\pi}{3}\)
2) \(\frac{\pi}{6}\)
3) \(\cos ^{-1}\left(\frac{5}{6}\right)\)
4) \(\cos ^{-1}\left(\frac{3}{4}\right)\)
Solution:
1) \(\frac{\pi}{3}\)
Dr’s of the lines are (a1, b1, c1) = (2, -1, 1); (a2, b2, c2) = (1, 1, 2)
∴ cos θ = \(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}=\frac{|2-1+2|}{\sqrt{4+1+1} \sqrt{1+1+4}}=\frac{3}{\sqrt{6} \cdot \sqrt{6}}=\frac{3}{6}=\frac{1}{2}=\cos 60^{\circ} \Rightarrow \theta=\frac{\pi}{3}\)
Question 7.
If θ is the acute angle between the lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})\) and \(\vec{r}=(2 \hat{i}-3 \hat{j}-4 \hat{k})+\mu(4 \hat{i}+3 \hat{j}+12 \hat{k})\) then cosθ
1) \(\frac{34}{39}\)
2) \(\frac{22}{39}\)
3) \(\frac{26}{39}\)
4) \(\frac{14}{39}\)
Solution:
4) \(\frac{14}{39}\)
Dr’s of the lines are (a1, b1, c1) = (1, 2, -2); (a2, b2, c2) = (4, 3, 12)
cos θ = \(\frac{|(4+6-24)|}{\sqrt{1+4+4} \sqrt{16+9+144}}=\frac{14}{3 \sqrt{169}}=\frac{14}{3(13)}=\frac{14}{39}\)
Question 8.
Equation of the line passing through (2, 1, -4) and parallel to the line joining the points (1, 0, -1) and (3, 2, 2) is
1) \(\frac{x+1}{2}=\frac{y+1}{2}=\frac{z-4}{3}\)
2) \(\frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)
3) \(\frac{x-2}{-2}=\frac{y-1}{-2}=\frac{z+4}{3}\)
4) \(\frac{x+2}{-2}=\frac{y+1}{-2}=\frac{z-4}{3}\)
Solution:
2) \(\frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)
Dr’s of line joning points (1, 0, -1) and (3, 2, 2) are (3 – 1, 2 – 0, 2 + 1) = (2, 2, 3)
required line || to given line ⇒ Dr’s of the line = (a, b, c) = (2, 2, 3)
Also (x1, y1, z1) = (2, 1, -4) is a point on the line
∴ Equation of required line = \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \Rightarrow \frac{x-2}{2}=\frac{y-1}{2}=\frac{z+4}{3}\)
Question 9.
Euation of the line passing through the point (1, 2, 3)and parallel to the z axis is
1) \(\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{0}\)
2) \(\frac{x-1}{0}=\frac{y-2}{1}=\frac{z-3}{1}\)
3) \(\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)
4) \(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\)
Solution:
3) \(\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)
Dr’s of z-axis (a, b, c) = (0, 0, 1) . Aslo point on the line is (x1, y1, z1) = (1, 2, 3)
∴ Equation of required line \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{x-1}{0}=\frac{y-2}{0}=\frac{z-3}{1}\)
Question 10.
The direction cossines of the line which is perpendicular to the lines \(\overline{\mathbf{r}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}})+\lambda(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\) and \(\stackrel{\rightharpoonup}{\mathbf{r}}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+\boldsymbol{\mu}(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) is
1) \(\left(\frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
2) \(\left(\frac{-3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{7}{\sqrt{59}}\right)\)
3) \(\left(\frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}}, \frac{7}{\sqrt{59}}\right)\)
4) \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
Solution:
4) \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)
Given lines \(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\mathrm{t} \overline{\mathrm{~b}} \Rightarrow \overline{\mathrm{~b}}=2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}}\) ………(1) and \(\overline{\mathrm{r}}=\overline{\mathrm{c}}+\mathrm{s} \overline{\mathrm{~d}} \Rightarrow \overline{\mathrm{~d}}=3 \overline{\mathrm{i}}-5 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) …………..(2)
Dr’s of the line which is perpendicular to both (1) and (2) and parallel to vector \(\overline{\mathbf{b}} \times \overline{\mathbf{d}}\)
Now \(\overline{\mathrm{b}} \times \overline{\mathrm{d}}=\left|\begin{array}{ccc}
\mathrm{i} & \mathrm{j} & \mathrm{k} \\
2 & -1 & 1 \\
3 & -5 & 2
\end{array}\right|=\overline{\mathrm{i}}(3)-\overline{\mathrm{j}}(1)+\overline{\mathrm{k}}(-7)\)
Dr’s of the line = (a, b, c) = (3, -1, -7) = \(\sqrt{3^2+(-1)^2+(-7)^2}=\sqrt{59}\)
∴ d.c’s = \(\left(\frac{3}{\sqrt{59}}, \frac{-1}{\sqrt{59}}, \frac{-7}{\sqrt{59}}\right)\)