Miscellaneous Exercise Practice
Chapter 11: Three Dimensional Geometry – NCERT Solutions
Q1
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Show that the line joining the origin to the point $(2, 1, 1)$ is perpendicular to the line determined by the points $(3, 5, -1), (4, 3, -1)$.
Line 1 (OA): Passes through $(0,0,0)$ and $(2,1,1)$. DRs are $(2-0, 1-0, 1-0) = (2, 1, 1)$.
Line 2 (BC): Passes through $(3,5,-1)$ and $(4,3,-1)$. DRs are $(4-3, 3-5, -1-(-1)) = (1, -2, 0)$.
Condition for perpendicularity: $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
$2(1) + 1(-2) + 1(0) = 2 - 2 + 0 = 0$.
$\boxed{\text{Lines are perpendicular}}$
Q2
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If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are $m_1n_2 - m_2n_1, n_1l_2 - n_2l_1, l_1m_2 - l_2m_1$.
Let the lines be represented by unit vectors $\vec{a} = l_1i + m_1j + n_1k$ and $\vec{b} = l_2i + m_2j + n_2k$.
Since lines are perpendicular, $\vec{a} \cdot \vec{b} = 0$.
A line perpendicular to both is parallel to $\vec{a} \times \vec{b}$.
$\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = i(m_1n_2 - m_2n_1) - j(l_1n_2 - l_2n_1) + k(l_1m_2 - l_2m_1)$.
Magnitude $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin 90^\circ = 1 \cdot 1 \cdot 1 = 1$.
Thus, the components are the direction cosines.
$\boxed{m_1n_2 - m_2n_1, n_1l_2 - n_2l_1, l_1m_2 - l_2m_1}$
Q3
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Find the angle between the lines whose direction ratios are $a, b, c$ and $b-c, c-a, a-b$.
Let vectors be $\vec{u} = ai + bj + ck$ and $\vec{v} = (b-c)i + (c-a)j + (a-b)k$.
$\vec{u} \cdot \vec{v} = a(b-c) + b(c-a) + c(a-b)$.
$= ab - ac + bc - ba + ca - cb = 0$.
Since dot product is 0, the angle is $90^\circ$.
$\boxed{90^\circ}$
Q4
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Find the equation of a line parallel to x-axis and passing through the origin.
Line passes through origin $(0,0,0)$.
Parallel to x-axis means direction ratios are proportional to $(1, 0, 0)$.
Equation: $\frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0}$.
$\boxed{\vec{r} = \lambda i \text{ or } y=0, z=0}$
Q5
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If the coordinates of the points A, B, C, D be $(1, 2, 3), (4, 5, 7), (-4, 3, -6)$ and $(2, 9, 2)$ respectively, then find the angle between the lines AB and CD.
DRs of AB: $(4-1, 5-2, 7-3) = (3, 3, 4)$.
DRs of CD: $(2-(-4), 9-3, 2-(-6)) = (6, 6, 8)$.
Ratios: $\frac{6}{3} = 2, \frac{6}{3} = 2, \frac{8}{4} = 2$.
Since direction ratios are proportional, the lines are parallel. Angle is $0^\circ$.
$\boxed{0^\circ}$
Q6
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If the lines $\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$ are perpendicular, find the value of k.
DRs of Line 1: $(-3, 2k, 2)$. DRs of Line 2: $(3k, 1, -5)$.
Perpendicular condition: $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$.
$-9k + 2k - 10 = 0 \Rightarrow -7k = 10$.
$\boxed{k = -\frac{10}{7}}$