Maths Case study (4 Marks)

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

2. (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

$=\begin{vmatrix}\frac{18-10+9}{\\\sqrt{3^2+(-2)^2+1^2}}\\\end{vmatrix}=\frac{17}{\sqrt{14}}\text{units}$

3. (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

$\frac{\text{x}-6}{3}=\frac{\text{y}-5}{-2}=\frac{\text{z}-9}{1}$

4. (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., $\Big(\frac{33}{14},\frac{104}{14},\frac{109}{14}\Big)$

5. (b) $\frac{3}{2}\sqrt{14}\text{sq}.\text{units}$

​​​​​​​Solution:

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

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

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

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

$=\frac{3}{2}\sqrt{14}\text{sq}.\text{units}$

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

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

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

$\Rightarrow\frac{\text{x}}{-4}=\frac{\text{y}}{3}=\frac{\text{z}-24}{12}$

2. (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., $-8\hat{\text{i}}-6\hat{\text{j}}-24\hat{\text{k}}$

3. (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

4. (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

5. (c) $-96\hat{\text{k}}$

Solution:

Sum of all vectors along the cables

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

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

$=-96\hat{\text{k}}$

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.

2. (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 $|\vec{\text{n}}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}$

3. (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

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

$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{(-6)}+\frac{\text{z}}{6}=1$

4. (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.

5. (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 $\sqrt{6}\text{ units}.$

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.

2. (c) 2 units

Solution:

Required length = Perpendicular distance from (3, -2, 1) to the plane x - 2y + 2z = 3

$=\begin{vmatrix}\frac{3-2(-2)+2(1)-3}{\\\sqrt{1^2+(-2)^2+2^2}}\\\end{vmatrix}=\frac{6}{3}=2\text{ units}$

3. (b) $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$

Solution:

The equation of perpendicular from the point (x₁, y₁, z₁) 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, x₁= 3, y₁= -2, z₁ = 1 and a = 1, b = -2, c = 2

$\therefore$ Required equation is $\frac{\text{x}-3}{1}=\frac{\text{y}+2}{-2}=\frac{\text{z}-1}{2}$

4. (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

5. (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 $\Big(\frac{5}{3},\frac{2}{3},\frac{-5}{3}\Big).$

1. (a) l₁l₂ + m₁m₂ + n₁n₂ = 0

Solution:

Since D.R'.s are proportional to D.C'.s, therefore L₁ will be perpendicular to L₂ iff

l₁l₂ + m₁m₂ + n₁n₂ = 0

2. (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 L₁ will be parallel to L₂, iff

$\frac{\text{l}_1}{\text{l}_2}=\frac{\text{m}_1}{\text{m}_2}=\frac{\text{n}_1}{\text{n}_2}$

3. (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).

4. (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 >.

5. (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.

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

2. (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., $<\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}}>$

3. (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 >.

4. (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.

5. (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).

1. (b) < 1, 0, 0 >

​​​​​​​Solution:

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

i.e., < 1, 0, 0 >

2. (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., $\frac{\text{x}}{1}=\frac{\text{y}}{2}=\frac{\text{z}}{3}$

3. (c) z = 0

​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

OABC is xy-plane, therefore its equation is z = 0.

4. (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.

5. (a) x = 1

​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Plane ABB' A' is parallel to yz-plane passing through ( 1, 0, 0), therefore its equation is x = 1.

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

2. (d) $\sqrt{11}\text{m}$

​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Required distance = Distance between P and Q

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

$=\sqrt{1+1+9}=\sqrt{11}\text{m}$

3. (d) $\frac{\sqrt{11}}{30}\text{seconds}$

​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

We know, Distance = Speed × Time

$\therefore$ Required time $=\frac{\sqrt{11}}{30}\text{seconds}.$

4. (b) $\frac{\text{x}-1}{1}=\frac{\text{y}-4}{-1}=\frac{\text{z}-2}{3}$

​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Equation of line PQ is $\frac{\text{x}-1}{1}=\frac{\text{y}-4}{-1}=\frac{\text{z}-2}{3}$

5. (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., $\Big(\frac{3}{4},\frac{3}{2},\frac{5}{4}\Big).$

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

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

$\Rightarrow\frac{\text{x}}{4}=\frac{\text{y}}{5}=\frac{30-\text{z}}{15}$

2. (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

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

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

$=\sqrt{1525}=\sqrt{61}\text{m}.$

3. (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., $-6\hat{\text{i}}+4\hat{\text{j}}-30\hat{\text{k}}$

4. (b) $17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$

​​​​​​​Solution:

Required sum

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

$17\hat{\text{i}}-6\hat{\text{j}}-90\hat{\text{k}}$

5. (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 $\text{OC}=\sqrt{15^2+20^2}=\sqrt{225+400}=\sqrt{625}$

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.

2. (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.

3. (b) 60km

​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

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

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

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

$\sqrt{3600}=60\text{km}.$

4. (d) 6km

​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Clearly, height of rocket from the ground = z - coordinate of given position = 6km.

5. (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., $\Big(\frac{-20}{13},\frac{-15}{13},\frac{-18}{13}\Big).$