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Maths Case study (4 Marks)

Three Dimensional Geometry · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 10 questions.

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Case study (4 Marks)

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10 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks
Suppose the floor of a hotel is made up of mirror polished Kota stone. Also, there is a large crystal chandelier attached at the ceiling of the hotel. Consider the floor of the hotel as a plane having equation $x - 2y + 2z = 3$ and crystal chandelier at the point $(3, -2, 1)$.

Based on the above information, answer the following questions.
  1. The d.r'.s of the perpendicular from the point $(3, -2, 1)$ to the plane $x - 2y + 2z = 3,$ is:
  1. $< 1, 2, 2 >$
  2. $< 1, -2, 2 >$
  3. $< 2, 1, 2 >$
  4. $< 2, -1, 2 >$
  1. The length of the perpendicular from the point $(3, -2, 1)$ to the plane $x - 2y + 2z = 3,$ is:
  1. $\frac{2}{3}\text{units}$
  2. $3$ units
  3. $2$ units
  4. None of these
  1. The equation of the perpendicular from the point $(3, -2, 1)$ to the plane $x - 2y + 2z = 3$, is:
  1. $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{-2}=\frac{\text{z}-1}{2}$
  2. $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$
  3. $\frac{\text{x}+3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$
  4. None of these
  1. The equation of plane parallel to the plane $x - 2y + 2z = 3,$ which is at a unit distance from the point $(3, -2, 1)$ is:
  1. $x - 2y + 2z = 0$
  2. $x - 2y + 2z = 6$
  3. $x - 2y + 2z = 12$
  4. Both $(b)$ and $(c)$
  1. The image of the point $(3, -2, 1)$ in the given plane is:
  1. $\Big(\frac{5}{3},\frac{2}{3},\frac{-5}{3}\Big)$
  2. $\Big(\frac{-5}{3},\frac{-2}{3},\frac{5}{3}\Big)$
  3. $\Big(\frac{-5}{3},\frac{2}{3},\frac{5}{3}\Big)$
  4. None of these
Answer
  1. (b) $< 1, -2, 2 >$
Solution:
Equation of plane is $x - 2y + 2z = 3$
$\therefore$ D.R.'s of normal to the plane are , which is also the D.R.'s of perpendicular from the point $(3, -2, 1)$ to the given plane.
  1. (c) $2$ units
Solution:
Required length = Perpendicular distance from $(3, -2, 1)$ to the plane $x - 2y + 2z = 3$
$=\begin{vmatrix}\frac{3-2(-2)+2(1)-3}{\\\sqrt{1^2+(-2)^2+2^2}}\\\end{vmatrix}=\frac{6}{3}=2\text{ units}$
  1. (b) $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$
Solution:
The equation of perpendicular from the point $(x_1, y_1, z_1)$ to the plane $ax + by + cz = d$ is given by
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
Here, $x_1= 3, y_1= -2, z_1 = 1$ and $a = 1, b = -2, c = 2$
$\therefore$ Required equation is $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$
  1. (d) Both $(b)$ and $(c)$
Solution:
The equation of the plane parallel to the plane $\text{x}-2\text{y}+2\text{z}-3=0$ is $\text{x}-2\text{y}+2\text{z}+\lambda=0$
Now, distance of this plane from the point $(3, -2, 1)$ is
$=\begin{vmatrix}\frac{3+4+2+\lambda}{\\\sqrt{1^2+(-2)^2+2^2}}\\\end{vmatrix}=\begin{vmatrix}\frac{9+\lambda}{3}\\\end{vmatrix}$
But, this distance is given to be unity
$\therefore|9+\lambda|=3$
$\Rightarrow\lambda+9=\pm3\Rightarrow\lambda=-6$
Or $-12$
Thus, required equation of planes are
$x - 2y + 2z- 6 = 0$ or $x - 2y + 2z - 12 = 0$
  1. (a) $\Big(\frac{5}{3},\frac{2}{3},\frac{-5}{3}\Big)$
Solution:
Let the coordinate of image of $(3, -2, 1)$ be
$Q(r + 3, -2r - 2, 2r + 1)$
Let R be the mid-point of $PQ$, then coordinate of R be
$\Big(\frac{\text{r}+6}{2},\frac{-2\text{r}-4}{2},\text{r}+1\Big)$
Since, R lies on the plane $x - 2y + 2z = 3$
$\therefore\Big(\frac{\text{r}+6}{2}\Big)-2\Big(\frac{-2\text{r}-4}{2}\Big)+2(\text{r}+1)=3$
$\Rightarrow9\text{r}=-12\Rightarrow\text{r}=-\frac{4}{3}$
Thus, the coordinates of $Q$ be $\Big(\frac{5}{3},\frac{2}{3},\frac{-5}{3}\Big).$
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Question 24 Marks
Consider the following diagram, where the forces in the cable are given. Based on the above information, answer the following questions.
  1. The equation of line along the cable AD is:
  1. $\frac{\text{x}}{5}=\frac{\text{y}}{4}=\frac{\text{z}-30}{15}$
  2. $\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{\text{z}-30}{15}$
  3. $\frac{\text{x}}{5}=\frac{\text{y}}{4}=\frac{30-\text{z}}{15}$
  4. $\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{30-\text{z}}{15}$
  1. The length of cable DC is:
  1. $4\sqrt{61}\text{m}$
  2. $5\sqrt{61}\text{m}$
  3. $6\sqrt{61}\text{m}$
  4. $7\sqrt{61}\text{m}$
  1. The vector DB is:
  1. $-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}}$
  2. $6\hat{\text{i}}-4\hat{\text{j}}+30\hat{\text{k}}$
  3. $6\hat{\text{i}}+4\hat{\text{j}}+30\hat{\text{k}}$
  4. None of these
  1. The sum of vectors along the cables is:
  1. $17\hat{\text{i}}+6\hat{\text{j}}+90\hat{\text{k}}$
  2. $17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$
  3. $17\hat{\text{i}}+6\hat{\text{j}}-90\hat{\text{k}}$
  4. None of these
  1. The sum of distances of points A, B and C from the origin, i.e., OA + OB + OC is:
  1. $\sqrt{164}+\sqrt{52}+\sqrt{625}$
  2. $\sqrt{52}+\sqrt{625}+\sqrt{48}$
  3. $\sqrt{164}+\sqrt{625}+\sqrt{49}$
  4. None of these
Answer
  1. (d) $\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{30-\text{z}}{15}$
Solution:

Clearly, the coordinates of A are (8, 10, 0) and D are (0, 0, 30)

$\therefore$ Equation of AD is given by

$\frac{\text{x}-0}{8-0}=\frac{\text{y}-0}{10-0}=\frac{30-\text{z}}{-30}$

$\Rightarrow\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{30-\text{z}}{15}$
  1. (b) $5\sqrt{61}\text{m}$
Solution:

The coordinates of point Care (15, -20, 0) and D are (0, 0, 30)

$\therefore$ Length of the cable DC

$=\sqrt{(0-15)^2+(0-20)^2+(30-0)^2}$

$=\sqrt{225+400+900}$

$=\sqrt{1525}=\sqrt{61}\text{m}.$
  1. (a) $-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}}$
Solution:

Since, the coordinates of point B are (-6, 4, 0) and D are (0, 0, 30), therefore vector DB is

$(-6-0)\hat{\text{i}}+(4-0)\hat{\text{j}}+(0-30)\hat{\text{k}},$

i.e., $-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}}$
  1. (b) $17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$
​​​​​​​Solution:

Required sum

$(8\hat{\text{i}}+10\hat{\text{j}}-30\hat{\text{k}})+(-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}})+(15\hat{\text{i}}-20\hat{\text{j}}-30\hat{\text{k}})$

$17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$
  1. (a) $\sqrt{164}+\sqrt{52}+\sqrt{625}$
​​​​​​​Solution:

Clearly, $\text{OA}=\sqrt{8^2+10^2}=\sqrt{164}$

$\text{OB}=\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}$

And $\text{OC}=\sqrt{15^2+20^2}=\sqrt{225+400}=\sqrt{625}$
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Question 34 Marks
Consider the following diagram, where the forces in the cable are given. Based on the above information, answer the following questions.
  1. The cartesian equation of line along EA is:
  1. $\frac{\text{x}}{-4}=\frac{\text{y}}{3}=\frac{\text{z}}{12}$
  2. $\frac{\text{x}}{-4}=\frac{\text{y}}{3}=\frac{\text{z}-24}{12}$
  3. $\frac{\text{x}}{-3}=\frac{\text{y}}{3}=\frac{\text{z}-12}{12}$
  4. $\frac{\text{x}}{3}=\frac{\text{y}}{4}=\frac{\text{z}-24}{12}$
  1. The vector $\overline{\text{ED}}$ is:
  1. $8\hat{\text{i}}-6\hat{\text{j}}+24\hat{\text{k}}$
  2. $-8\hat{\text{i}}-6\hat{\text{j}}+24\hat{\text{k}}$
  3. $-8\hat{\text{i}}-6\hat{\text{j}}-24\hat{\text{k}}$
  4. $8\hat{\text{i}}+6\hat{\text{j}}+24\hat{\text{k}}$
  1. The length of the cable EB is:
  1. 24 units
  2. 26 units
  3. 27 units
  4. 25 units
  1. The length of cable EC is equal to the length of:
  1. EA
  2. EB
  3. ED
  4. All of these
  1. The sum of all vectors along the cables is:
  1. $96\hat{\text{i}}$
  2. $96\hat{\text{j}}$
  3. $-96\hat{\text{k}}$
  4. $96\hat{\text{k}}$
Answer
  1. (b) $\frac{\text{x}}{-4}=\frac{\text{y}}{3}=\frac{\text{z}-24}{12}$
​​​​​​​Solution:

Clearly, the coordinates of A are (8, -6, 0) and that of E are (0, 0, 24).

Also, cartesian equation of line along EA is given by

$\frac{\text{x}-0}{8-0}=\frac{\text{y}-0}{-6-0}=\frac{\text{z}-24}{0-24}$

$\Rightarrow\frac{\text{x}}{8}=\frac{\text{y}}{-6}=\frac{\text{z}-24}{-24}$

$\Rightarrow\frac{\text{x}}{-4}=\frac{\text{y}}{3}=\frac{\text{z}-24}{12}$
  1. (c) $-8\hat{\text{i}}-6\hat{\text{j}}-24\hat{\text{k}}$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Clearly, the coordinates of D are (-8, -6, 0) and that of E are (0, 0, 24)

$\therefore$ Vector $\overline{\text{ED}}$ is $ (-8-0)\hat{\text{i}}+(-6-0)\hat{\text{j}}+(0-24)\hat{\text{k}},$

i.e., $-8\hat{\text{i}}-6\hat{\text{j}}-24\hat{\text{k}}$
  1. (b) 26 units
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Since, the coordinates of B are (8, 6, 0) and that of E are (0, 0, 24), therefore length of cable

$\text{EB}=\sqrt{(8-0)^2+(6-0)^2+(0-24)^2}$

$=\sqrt{64+36+576}=\sqrt{676}$

26 units
  1. (d) All of these
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Since the coordinates of C are (-8, 6, 0) therefore length of cable

$\text{EC}=\sqrt{(-8-0)^2+(6-0)^2+(0-24)^2}$

$=\sqrt{64+36+576}=\sqrt{676}$

26 units

Similarly, length of cable EA = ED = 26 units
  1. (c) $-96\hat{\text{k}}$
Solution:

Sum of all vectors along the cables

$=\overline{\text{EA}}+\overline{\text{EB}}+\overline{\text{EC}}+\overline{\text{ED}}$

$= (8\hat{\text{i}}-6\hat{\text{j}}-24\hat{\text{k}})+(8\hat{\text{i}}+6\hat{\text{j}}-24\hat{\text{k}})+(-8\hat{\text{i}}+6\hat{\text{j}}-24\hat{\text{k}})+(-8\hat{\text{i}}-6\hat{\text{j}}-24\hat{\text{k}})$

$=-96\hat{\text{k}}$
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Question 44 Marks
If $a_{1,} b_{1,} c_{1,}$ and $a_{2,}b_{2,} c_2$ are direction ratios of two lines say $L_1$ and $L_2$ respectively. Then $L_1 || L_2$ iff $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ and $\text{L}_1\perp\text{L}_2$ iff $a_1a_2 + b_1b_2 + c_1c_2 = 0.$

 Based on the above information, answer the following questions.
  1. If $l_{1,} m_1, n_{1,}$ and $l_2, m_2, n_2$ are the direction cosines of $L_1$ and $L_2$ respectively, then $L_1$ will be perpendicular to $L_2$, iff:
  1. $l_1l_2 + m_1m_2 + n_1n_2 = 0$
  2. $l_1m_2 + m_1l_2 + n_1n_2 = 0$
  3. $\frac{\text{l}_1}{\text{l}_2}=\frac{\text{m}_1}{\text{m}_2}=\frac{\text{n}_1}{\text{n}_2}$
  4. None of these
  1. If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of $L_1$ and $L_2$ respectively, then $L_1$ will be parallel to $L_2,$ iff:
  1. $l_1l_2 + m_1m_2 + n_1n_2 = 0$
  2. $l_1m_2 + m_1l_2 + n_1n_2 = 0$
  3. $\frac{\text{l}_1}{\text{l}_2}=\frac{\text{m}_1}{\text{m}_2}=\frac{\text{n}_1}{\text{n}_2}$
  4. $m_1n_2 + m_2n_2 + l_1l_2 = 0$
  1. The coordinates of the foot of the perpendicular drawn from the point $A(1, 2, 1)$ to the line joining $B(1, 4, 6)$ and $C(5, 4, 4),$ are:
  1. $(1, 2, 1)$
  2. $(2, 4, 5)$
  3. $(3, 4, 5)$
  4. $(3, 4, 5)$
  1. The direction ratios of the line which is perpendicular to the lines with direction ratios proportional to $(1, -2, -2)$ and $(0, 2, 1)$ are:
  1. $< 1, 2, 1 >$
  2. $< 2,-1, 2 >$
  3. $< -1,2, 2 >$
  4. None of these
  1. The lines $\frac{\text{x}-2}{3}=\frac{\text{y}+1}{-2}=\frac{\text{z}-2}{0}$ and $\frac{\text{x}-1}{1}=\frac{\frac{\text{y}+3}{2}}{\frac{3}{2}}=\frac{\text{z}+5}{2}$ are:
  1. Parallel.
  2. Perpendicular.
  3. Skew lines.
  4. Non-intersecting.
Answer
  1. (a) $l_1l_2 + m_1m_2 + n_1n_2 = 0$
Solution:
Since D.R'.s are proportional to D.C'.s, therefore $L_1$ will be perpendicular to $L_2$ iff
$l_1l_2 + m_1m_2 + n_1n_2 = 0$
  1. (c) $\frac{\text{l}_1}{\text{l}_2}=\frac{\text{m}_1}{\text{m}_2}=\frac{\text{n}_1}{\text{n}_2}$
Solution:
Since D.R'.s are proportional to D.C'.s, therefore $L_1$ will be parallel to $L_2,$ iff
$\frac{\text{l}_1}{\text{l}_2}=\frac{\text{m}_1}{\text{m}_2}=\frac{\text{n}_1}{\text{n}_2}$
  1. (c) $(3, 4, 5)$
Solution:
Equation of line joining $B$ and $C$ is
$\frac{\text{x}-1}{4}=\frac{\text{y}-4}{0}=\frac{\text{z}-6}{-2}$
Let coordinates of foot of perpendicular be $D(x, y, z).$
$\therefore$ D.R'.s of AD are $< x - 1, y - 2, z - 1 >.$
Now, $4(x - 1) +0(y - 2) -2(z - 1) = 0$
$⇒ 4x - 2z = 2$
Also, $(x, y, z)$ will satisfy equation of line $BC$.
Here, $(3, 4, 5)$ satisfy both the conditions.
$\therefore$ Required coordinates are $(3, 4, 5).$
  1. (b) $< 2, -1, 2 >$
​​​​​​​​​​​​​​Solution:​​​​​​​
Let $a, b, c$ be the direction ratios of the required line. Since it is perpendicular to the lines whose direction ratios are $( 1, -2, -2)$ and $(0, 2, 1)$ respectively.
$\therefore a - 2b - 2c = 0 (i)$
$0. a + 2b + c = 0 (ii)$
On solving $(i)$ and $(ii)$ by cross-multiplication, we get
$\frac{\text{a}}{-2+4}=\frac{\text{b}}{0-1}=\frac{\text{c}}{2}$
$\Rightarrow\frac{\text{a}}{2}=\frac{\text{b}}{-1}=\frac{\text{c}}{2}$
Thus, the direction ratios of the required line are $< 2, -1, 2 >.$​​​​​​​
  1. (b) Perpendicular.
​​​​​​​​​​​​​​​​​​​​​Solution:
D.R'.s of given lines are $< 3, -2, 0 >$ and $<1,\frac{3}{2},2>$
Now, as $ 3.1+ (-2).\Big(\frac{3}{2}\Big)+0.2=3-3+0=0$
$\therefore$ Given lines are perpendicular to each other.
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Question 54 Marks
Two motorcycles A and Bare running at the speed more than allowed speed on the road along the lines $\vec{\text{r}}=\lambda(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$ and $\vec{\text{r}}=3\hat{\text{i}}+3\hat{\text{j}}+\mu(2\hat{\text{i}+\hat{\text{j}}+\hat{\text{k}}}),$ respectively. Based on the above information, answer the following questions.
  1. The cartesian equation of the line along which motorcycle A is running is:
  1. $\frac{\text{x}+1}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}-1}{-1}$
  2. $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{-1}$
  3. $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{1}$
  4. None of these
  1. The direction cosines of line along which motorcycle A is running, are:
  1. < 1, -2, 1 >
  2. < I, 2, -1 >
  3. $<\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}>$
  4. $<\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}>$
  1. The direction ratios of line along which motorcycle Bis running, are:
  1. < 1, 0, 2 >
  2. < 2, 1, 0 >
  3. < 1, 1, 2 >
  4. < 2, 1, 1 >
  1. The shortest distance between the gives lines is:
  1. 4 units
  2. $2\sqrt{3}\text{ units}$
  3. $3\sqrt{2}\text{ units}$
  4. 0 units
  1. The motorcycles will meet with an accident at the point:
  1. (-1, 1, 2)
  2. (2, 1, -1)
  3. (1, 2, -1)
  4. Does not exist
Answer
  1. (b) $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{-1}$
Solution:

The line along which motorcycle A is running is $\vec{\text{r}}=\lambda(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}),$ which can be rewritten as $(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}-\text{z}\hat{\text{k}})=\lambda\hat{\text{i}}+2\lambda\hat{\text{j}}-\lambda\hat{\text{k}}$

$\Rightarrow\text{x}=\lambda,\text{ y}=2\lambda,\text{ z}=-\lambda$

$\Rightarrow\frac{\text{x}}{1}=\lambda,\frac{\text{y}}{2}=\lambda,\frac{\text{x}}{-1}=\lambda$

Thus, the required cartesian equation is $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{-1}$
  1. (d) $<\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}>$
Solution:

Clearly, DR'.s of the required line are < 1, 2, -1 >

$\therefore$ D.C'.s are

$<\frac{1}{\sqrt{1^2+2^2+(-1)^2}},\frac{2}{\sqrt{1^2+2^2+(-1)^2}},\frac{-1}{\sqrt{1^2+2^2+(-1)^2}}>$

i.e., $<\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}>$
  1. (d) < 2, 1, 1 >
Solution:

The line along which motorcycle Bis running is $\vec{\text{r}}=(3\hat{\text{i}}+3\hat{\text{j}})+\mu(2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}),$ which is parallel to the vector $2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}.$

$\therefore$ D.R.'s of the required line are < 2, 1, 1 >.
  1. (d) 0 units
​​​​​​​Solution:

Here, $\vec{\text{a}}_1=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}},$

$\vec{\text{a}}_2=3\hat{\text{i}}+3\hat{\text{j}},$

$\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}_2=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},$

$\therefore\vec{\text{a}}_1-\vec{\text{a}}_2=3\hat{\text{i}}+3\hat{\text{j}}$

And $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&-1\\2&1&1\end{vmatrix}$

$=3\hat{\text{i}}-3\hat{\text{j}}-3\hat{\text{k}}$

Now, $(\vec{\text{a}}_1-\vec{\text{a}}_2)\cdot(\vec{\text{b}}_1\times\vec{\text{b}}_2)$

$=(3\hat{\text{i}}+3\hat{\text{j}})\cdot(3\hat{\text{i}}-3\hat{\text{j}}-3\hat{\text{k}})$

= 9 - 9 = 0

Hence, shortest distance between the given lines is 0.
  1. (c) (1, 2, -1)
​​​​​​​​​​​​​​​​​​​​​Solution:

Since, the point (1, 2, -1) satisfy both the equations of lines, therefore point of intersection of given lines is (1, 2, -1). So, the motorcycles will meet with an accident at the point (1, 2, -1).
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Question 64 Marks
In a diamond exhibition, a diamond is covered in cubical glass box having coordinates O(0, 0, 0), A(1, 0, 0), B(1, 2, 0), C(0, 2, 0), O'(0, 0, 3), A'(1, 0, 3), B'(1, 2, 3) and C'(0, 2, 3). Based on the above information, answer the following questions.
  1. Direction ratios of OA are:
  1. < 0, 1, 0 >
  2. < 1, 0, 0 >
  3. < 0, 0, 1 >
  4. None of these
  1. Equation of diagonal OB' is:
  1. $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$
  2. $\frac{\text{x}}{0}=\frac{\text{y}}{1}=\frac{\text{z}}{2}$
  3. $\frac{\text{x}}{1}=\frac{\text{y}}{0}=\frac{\text{z}}{2}$
  4. None of these
  1. Equation of plane OABC is:
  1. x = 0
  2. y = 0
  3. z = 0
  4. None of these
  1. Equation of plane O' A' B' C' is:
  1. x = 3
  2. y = 3
  3. z = 3
  4. z = 2
  1. Equation of plane ABB' A' is:
  1. x = 1
  2. y = 1
  3. z = 2
  4. x = 3
Answer
  1. (b) < 1, 0, 0 >
​​​​​​​Solution:

D.R'. s of OA are < 1 - 0, 0 - 0, 0 - 0 >,

i.e., < 1, 0, 0 >
  1. (a) $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$
​​​​​​​​​​​​​​​​​​​​​Solution:

Equation of diagonal OB' is

$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{2}=\frac{\text{z}-0}{3}$

i.e., $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$
  1. (c) z = 0
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

OABC is xy-plane, therefore its equation is z = 0.
  1. (c) z = 3
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Plane O' A' B' C' is parallel to xy-plane passing through (0, 0, 3), therefore its equation is z = 3.
  1. (a) x = 1
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Plane ABB' A' is parallel to yz-plane passing through ( 1, 0, 0), therefore its equation is x = 1.
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Question 74 Marks
The Indian Coast Guard (ICG) while patrolling, saw a suspicious boat with four men. They were nowhere looking like fishermen. The soldiers were closely observing the movement of the boat for an opportunity to seize the boat. They observe that the boat is moving along a planar surface. At an instant of time, the coordinates of the position of coast guard helicopter and boat are (2, 3, 5) and (1, 4, 2) respectively. Based on the above information, answer the following questions.
  1. If the line joining the positions of the helicopter and boat is perpendicular to the plane in which boat moves, then equation of plane is:
  1. x - y + 3z = 2
  2. x + y + 3z = 2
  3. x - y + 3z = 3
  4. x + y + 3z = 3
  1. If the soldier decides to shoot the boat at given instant of time, where the distance measured in metres then what is the distance that bullet has to travel?
  1. $\sqrt{5}\text{m}$
  2. $\sqrt{8}\text{m}$
  3. $\sqrt{10}\text{m}$
  4. $\sqrt{11}\text{m}$
  1. If the speed of bullet is 30m/ sec, then how much time will the bullet take to hit the boat after the shot is fired?
  1. 30 seconds
  2. 1 second
  3. $\frac{1}{2}\text{second}$
  4. $\frac{\sqrt{11}}{30}\text{seconds}$
  1. At the given instant of time, the equation of line passing through the positions of helicopter and boat is:
  1. $\frac{\text{x}}{1}=\frac{\text{y}}{-1}=\frac{\text{z}}{3}$
  2. $\frac{\text{x}-1}{1}=\frac{\text{y}-4}{-1}=\frac{\text{z}-2}{3}$
  3. $\frac{\text{x}}{1}=\frac{\text{y}}{1}=\frac{\text{z}}{-3}$
  4. $\frac{\text{x}-1}{1}=\frac{\text{y}-4}{1}=\frac{\text{z}-2}{-3}$
  1. At a different instant of time, the boat moves to a different position along the planar surface. What should be the coordinates of the location of the boat for the bullet to hit the boat if soldier shoots the bullet along the line whose equation is $\frac{\text{x}-1}{1}=\frac{\text{y}-1}{-2}=\frac{\text{z}-2}{3}?$
  1. $\Big(\frac{1}{2},\frac{1}{2},\frac{1}{2}\Big)$
  2. $\Big(\frac{3}{4},\frac{3}{2},\frac{5}{4}\Big)$
  3. $\Big(\frac{1}{3},\frac{1}{4},\frac{1}{5}\Big)$
  4. None of these
Answer
  1. (c) x - y + 3z = 3
​​​​​​​​​​​​​​Solution:

Let P(2, 3, 5) and Q( 1, 4, 2) be the positions of helicopter and boat respectively.

Now, direction ratios of PQ are proportional to 1 - 2, 4 - 3, 2 - 5, i.e., - 1, 1, - 3.

So, equation of plane passing through Q( 1, 4, 2) and perpendicular to PQ is

-(x - 1) + (y - 4) + (-3) (z - 2) = 0 ⇒ x - y + 3z = 3
  1. (d) $\sqrt{11}\text{m}$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Required distance = Distance between P and Q

$=\sqrt{(1-2)^2+(4-3)^2+(2-5)^2}$

$=\sqrt{1+1+9}=\sqrt{11}\text{m}$
  1. (d) $\frac{\sqrt{11}}{30}\text{seconds}$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

We know, Distance = Speed × Time

$\therefore$ Required time $=\frac{\sqrt{11}}{30}\text{seconds}.$
  1. (b) $\frac{\text{x}-1}{1}=\frac{\text{y}-4}{-1}=\frac{\text{z}-2}{3}$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Equation of line PQ is $\frac{\text{x}-1}{1}=\frac{\text{y}-4}{-1}=\frac{\text{z}-2}{3}$
  1. (b) $\Big(\frac{3}{4},\frac{3}{2},\frac{5}{4}\Big)$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Any point on the line $\frac{\text{x}-1}{1}=\frac{\text{y}-1}{-2}=\frac{\text{z}-2}{3}$ is given by

$(\lambda+1, -2\lambda+1,3\lambda+2).$

Now, on substituting this point in the equation of plane x - y + 3z = 3, we get

$(\lambda+1)-(-2\lambda+1)+3(3\lambda+2)=3$

$\Rightarrow\lambda+1+2\lambda-1+9\lambda+6=3$​​​​​​​

$\Rightarrow12\lambda=-3$

$\Rightarrow\lambda=\frac{-1}{4}$

Thus, the required point is $\Big(\frac{-1}{4}+1,\frac{1}{2}+1,\frac{-3}{4}+2\Big)$

i.e., $\Big(\frac{3}{4},\frac{3}{2},\frac{5}{4}\Big).$
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Question 84 Marks
The equation of motion of a rocket are: x = 2t, y = -4t, z = 41, where the time 't' is given in seconds, and the distance measured is in kilometres. Based on the above information, answer the following questions.
  1. What is the path of the rocket?
  1. Straight line.
  2. Circle.
  3. Parabola.
  4. None of these.
  1. Which of the following points lie on the path of the rocket?
  1. (0, 1, 2)
  2. (1, -2, 2)
  3. (2, -2, 2)
  4. None of these
  1. At what distance will the rocket be from the starting point (0, 0, 0) in 10 seconds?
  1. 40km
  2. 60km
  3. 30km
  4. 80km
  1. If the position of rocket at certain instant of time is (3, -6, 6), then what will be the height of the rocket from the ground, which is along the xy-plane?
  1. 3km
  2. 2km
  3. 4km
  4. 6km
  1. At certain instant of time, if the rocket is above sea level, where equation of surface of sea is given by 3x - y + 4z = 2 and position of rocket at that instant of time is (1, -2, 2), then the image of position of rocket in the sea is:
  1. $\Big(\frac{20}{13},\frac{15}{13},\frac{18}{13}\Big)$
  2. $\Big(\frac{-20}{13},\frac{-15}{13},\frac{-18}{13}\Big)$
  3. $\Big(\frac{20}{13},\frac{-15}{13},\frac{18}{13}\Big)$
  4. None of these
Answer
  1. (a) Straight line.
Solution:

Eliminating 't' from the given equations, we get equation of path as,

$\frac{\text{x}}{2}=\frac{\text{y}}{-4}=\frac{\text{z}}{4}$ or $\frac{\text{x}}{1}=\frac{\text{y}}{-2}=\frac{\text{z}}{2}.$

​​​​​​​Thus, the path of the rocket represents a straight line.
  1. (b) (1, -2, 2)
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Since, only (1, -2, 2) satisfy the equation of path of rocket, therefore ( 1, -2, 2) lie on the path of rocket.
  1. (b) 60km
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Fort = 10sec, we have x = 20, y = -40, z = 40

Now, required distance $\sqrt{\text{x}^2+\text{y}^2+\text{z}^2}$

$=\sqrt{20^2+(-40)^2+(40)^2}=\sqrt{400+1600+1600}$

$\sqrt{3600}=60\text{km}.$
  1. (d) 6km
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Clearly, height of rocket from the ground = z - coordinate of given position = 6km.
  1. (b) $\Big(\frac{-20}{13},\frac{-15}{13},\frac{-18}{13}\Big)$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Let Q be the image of point P(1, -2, 2) in the plane 3x - y + 4z = 2. Then, equation of PQ is

$\frac{\text{x}-1}{3}=\frac{\text{y}+2}{-1}=\frac{\text{z}-2}{4}$

Let the coordinates of Q be (3r + 1, -r - 2, 4r + 2). Let R be the mid-point of PQ. Then, coordinates of R are

$\Big(\frac{3\text{r}+2}{2},\frac{\text{-r}-4}{2},\frac{\text{4r}+4}{2}\Big)$ or $\Big(\frac{3}{2}\text{r}+1,\frac{\text{-r}}{2}-2,\text{2r}+2\Big)$

Since, R lies on 3x - y + 4z = 2.

$\therefore3\Big(\frac{3}{2}\text{r}+1\Big)-\Big(\frac{\text{-r}}{2}-2\Big)+4(\text{2r}+2)=2$

$\Rightarrow\frac{\text{9r}}{2}+3+\frac{\text{r}}{2}+2+\text{8r}+8=2$

$\Rightarrow\text{13r}+13-2$

$\Rightarrow\text{r}=\frac{-11}{13}$

Hence, the coordinates of Q are

$\Big(\frac{-33}{13}+1,\frac{11}{13}-2,\frac{-44}{13}+2\Big)$

i.e., $\Big(\frac{-20}{13},\frac{-15}{13},\frac{-18}{13}\Big).$
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Question 94 Marks
A mobile tower stands at the top of a hill. Consider the surface on which tower stand as a plane having points A(0, 1, 2), B(3, 4, -1), and C(2, 4, 2) on it. The mobile tower is tied with 3 cables from the point A, Band C such that it stand vertically on the ground. The peak of the tower is at the point ( 6, 5, 9), as shown in the figure. Based on the above information, answer the following questions.
  1. The equation of plane passing through the points A, Band C is:
  1. 3x - 4y + z = 0
  2. 3x - 2y + z = 0
  3. 3x - 2y + z = 0
  4. 4x - 3y + 3z = 0
  1. The height of the tower from the ground is:
  1. 6 units
  2. 5 units
  3. $\frac{17}{\sqrt{14}}\text{units}$
  4. $\frac{5}{\sqrt{14}}\text{units}$
  1. The equation of line of perpendicular drawn from the peak of tower to the ground is:
  1. $\frac{\text{x}-6}{3}=\frac{\text{y}-4}{-2}=\frac{\text{z}-9}{1}$
  2. $\frac{\text{x}-6}{3}=\frac{\text{y}-5}{-2}=\frac{\text{z}-9}{1}$
  3. $\frac{\text{x}-6}{3}=\frac{\text{y}-4}{2}=\frac{\text{z}-9}{1}$
  4. $\frac{\text{x}-6}{3}=\frac{\text{y}-5}{2}=\frac{\text{z}-9}{1}$
  1. The coordinates of foot of perpendicular drawn from the peak of tower to the ground are:
  1. $\Big(\frac{33}{14},\frac{104}{14},\frac{109}{14}\Big)$
  2. $\Big(\frac{33}{14},\frac{109}{14},\frac{104}{14}\Big)$
  3. $\Big(\frac{33}{14},\frac{105}{14},\frac{109}{14}\Big)$
  4. None of these
  1. The area of $\triangle\text{ABC}$ is:
  1. $\frac{1}{2}\sqrt{14}\text{sq}.\text{units}$
  2. $\frac{3}{2}\sqrt{14}\text{sq}.\text{units}$
  3. $\sqrt{14}\text{sq}.\text{units}$
  4. $2\sqrt{14}\text{sq}.\text{units}$
Answer
  1. (b) 3x - 2y + z = 0
Solution:

The equation of plane passing through three non-collinear points is given by

$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1\\\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{x}_3-\text{x}_1&\text{y}_3-\text{y}_1&\text{z}_3-\text{z}_1\end{vmatrix}=0$

$\Rightarrow\begin{vmatrix}\text{x}&\text{y}-1&\text{z}-2\\2-0&4-1&-1-2\\2-0&4-1&2-2\end{vmatrix}=0$

$\Rightarrow\begin{vmatrix}\text{x}&\text{y}-1&\text{z}-2\\3&3&-3\\2&3&0\end{vmatrix}=0$

⇒ x (0 + 9) - (y- 1) (0 + 6) + (z - 2) (9- 6) = 0

⇒ 9x - 6y + 6 + 3z - 6 = 0

⇒ 3x - 2y + z = 0
  1. (c) $\frac{17}{\sqrt{14}}\text{units}$
Solution:

Height of tower = Perpendicular distance from the point (6, 5, 9) to the plane 3x - 2y + z = 0

$=\begin{vmatrix}\frac{18-10+9}{\\\sqrt{3^2+(-2)^2+1^2}}\\\end{vmatrix}=\frac{17}{\sqrt{14}}\text{units}$
  1. (b) $\frac{\text{x}-6}{3}=\frac{\text{y}-5}{-2}=\frac{\text{z}-9}{1}$
Solution:

D.R.'s of perpendicular are< 3, -2, 1 >

[$\because$ Perpendicular is parallel to the normal to the plane] Since, perpendicular is passing through the point (6, 5, 9), therefore its equation is

$\frac{\text{x}-6}{3}=\frac{\text{y}-5}{-2}=\frac{\text{z}-9}{1}$
  1. (a) $\Big(\frac{33}{14},\frac{104}{14},\frac{109}{14}\Big)$
Solution:

Let the coordinates of foot of perpendicular are $(3\lambda+6,-2\lambda+5,\lambda+9)$

Since, this point lie on the plane 3x - 2y + z = 0, therefore we get

$\Rightarrow9\lambda+4\lambda+\lambda+18-10+9=0$

$\Rightarrow14\lambda=-17$

$\Rightarrow\lambda=\frac{-17}{14}$

Thus, the coordinates of foot of perpendicular are

$\Big(\frac{-51}{14}+6,\frac{34}{14}+5,\frac{-17}{14}+9\Big)$

i.e., $\Big(\frac{33}{14},\frac{104}{14},\frac{109}{14}\Big)$
  1. (b) $\frac{3}{2}\sqrt{14}\text{sq}.\text{units}$
​​​​​​​Solution:

Clearly, area of $\text{ABC}=\frac{1}{2}|\overline{\text{AB}}\times\overline{\text{AC}}|$

$=\frac{1}{2}|(3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})\times(2\hat{\text{i}}+3\hat{\text{k}})|$

$=\frac{1}{2}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&3&-3\\2&3&0\end{vmatrix}=\frac{1}{2}|9\hat{\text{i}}-6\hat{\text{j}}+3\hat{\text{k}}|$​​​​​​​

$=\frac{1}{2}\sqrt{9^2+6^2+3^2}=\frac{1}{2}\sqrt{126}$

$=\frac{3}{2}\sqrt{14}\text{sq}.\text{units}$
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Question 104 Marks
A football match is organised between students of class XII of two schools, say school A and school B. For which a team from each school is chosen. Remaining students of class XII of school A and Bare respectively sitting
on the plane represented by the equation$ \vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{2\text{k}})=5$ and $ \vec{\text{r}}.(\hat{2\text{i}}-\hat{\text{j}}+\hat{\text{k}})=6$ to cheer up the team of their respective schools.
Based on the above information, answer the following questions.
  1. The cartesian equation of the plane on which students of school A are seated is:
  1. 2x - y + z = 8
  2. 2x + y + z = 8
  3. x + y + 2z = 5
  4. x + y + z = 5
  1. The magnitude of the normal to the plane on which students of school Bare seated, is:
  1. $\sqrt{5}$
  2. $\sqrt{6}$
  3. $\sqrt{3}$
  4. $\sqrt{2}$
  1. The intercept form of the equation of the plane on which students of school Bare seated is:
  1. $\frac{\text{x}}{6}+\frac{\text{y}}{6}+\frac{\text{z}}{6}=1$
  2. $\frac{\text{x}}{3}+\frac{\text{y}}{(-6)}+\frac{\text{z}}{6}=1$
  3. $\frac{\text{x}}{3}+\frac{\text{y}}{6}+\frac{\text{z}}{6}=1$
  4. $\frac{\text{x}}{3}+\frac{\text{y}}{6}+\frac{\text{z}}{3}=1$
  1. Which of the following is a student of school B?
  1. Mohit sitting at (1, 2, 1)
  2. Ravi sitting at (0, 1, 2)
  3. Khushi sitting at (3, 1, 1)
  4. Shewta sitting at (2, -1, 2)
  1. The distance of the plane, on which students of school Bare seated, from the origin is:
  1. 6 units
  2. $\frac{1}{\sqrt{6}}\text{ units}$
  3. $\frac{5}{\sqrt{6}}\text{ units}$
  4. $\sqrt{6}\text{ units}$
Answer
  1. (c) x + y + 2z = 5
Solution:

Clearly, the plane for students of school A is $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{2\text{k}})=5 $ which can be rewritten as $(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{2\text{k}})=5$

⇒ x + y + 2z = 5, which is the required cartesian equation.
  1. (b) $\sqrt{6}$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Clearly, the equation of plane for students of school B is $\vec{\text{r}}.(\hat{2\text{i}}-\hat{\text{j}}+\hat{\text{k}})=6$ which is of the form $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$

$\therefore$ Normal vector to the plane is, $\vec{\text{n}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and

Its magnitude is $|\vec{\text{n}}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}$
  1. (b) $\frac{\text{x}}{3}+\frac{\text{y}}{(-6)}+\frac{\text{z}}{6}=1$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

The cartesian form is 2x - y + z = 6, which can be rewritten as

$\frac{2\text{x}}{6}-\frac{\text{y}}{6}+\frac{\text{z}}{6}=1$

$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{(-6)}+\frac{\text{z}}{6}=1$
  1. (c) Khushi sitting at (3, 1, 1)
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Since, only the point (3, 1, 1) satisfy the equation of plane representing seating position of students of school B, therefore Khushi is the student of school B.
  1. (d) $\sqrt{6}\text{ units}$
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Equation of plane representing students of school B is $\vec{\text{r}}.(\hat{2\text{i}}-\hat{\text{j}}+\hat{\text{k}})=6,$ which is not in normal form, as $|\vec{\text{n}}|\not=1.$

On dividing both sides by $\sqrt{2^2+(-1)^2+1^2}=\sqrt{6},$ we get

$\vec{\text{r}}.\Big(\frac{2}{\sqrt{6}}\hat{\text{i}}-\frac{1}{\sqrt{6}}\hat{\text{j}}+\frac{1}{\sqrt{6}}\hat{\text{k}}\Big)$

$=\frac{6}{\sqrt{6}},$

Which is of the form $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$

Thus, the required distance is $\sqrt{6}\text{ units}.$
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