Question 1515 Marks
Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.
AnswerWe know to find the equation pf plane that bisects A(1, 2, 3) and B(3, 4, 5) perpendicularly
We know that, equation of plane passing through the point $\vec{\text{a}}$ and perpendicular to vector $\vec{\text{n}}$ is given by,
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0\ ...(\text{i})$
Here, $\vec{\text{a}}=\text{mid-point of AB}$
$=\frac{\text{position vector of A}+\text{position vector of B}}{2}$
$=\frac{\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}}{2}$
$\vec{\text{a}}=\frac{4\hat{\text{i}}+6\hat{\text{j}}+8\hat{\text{k}}}{2}$
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
And, $\vec{\text{n}}=\overrightarrow{\text{AB}}$
= Position vector of B - Position vector of A
$=(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}})-(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
$=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{n}}=2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Put, the value of $\vec{\text{a}}$ and $\vec{\text{n}}$ in equation (i),
$\vec{\text{r}}-(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})-\big[(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})\big]=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})-[(2)(2)+(3)(2)+(4)(2)]=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})-[4+6+8]=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})-18=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})=18$
View full question & answer→Question 1525 Marks
If the line $\frac{\text{x}-3}{2}=\frac{\text{y}+2}{-1}=\frac{\text{z}+4}{3}$ lies in the plane $lx + my - z = 9$, then find the value of $l^2 + m^2.$
AnswerThe line $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$ lies in the plane $Ax + By + Cz + D = 0$ if $(i) Ax_1 + By_1 + Cz_1 + D = 0$ and (ii) $aA + bB + cC = 0$
It is given that the line $\frac{\text{x}-3}{2}=\frac{\text{y}+2}{-1}=\frac{\text{z}+4}{3}$ lies in the plane lx + my - z = 9
$\therefore$ $l \times 3 + m \times (-2) - (-4) = 9$
$\Rightarrow 3l - 2m = 5 ....(i)$
Also,
$2 \times l + (-1) \times m + 3 \times (-1) = 0$
$\Rightarrow 2l - m = 3 ....(ii)$
Solving (i) and (ii) we get
l = 1 and m = -1
$\therefore$ $l^2 + m^2 = 1^2+ (-1)^2 = 1 + 1 = 2$
Thus, the value of $l^2 + m^2$ is $2$
View full question & answer→Question 1535 Marks
The equation of tangent at (2, 3) on the curve $y^{2} = \text{ax}^{3} + \text{b is y = 4x - 5}. $ Find the value of a and b.
Answer$\text{y}^{2} = \text{ax}^{3} + \text{b} \Rightarrow\text{2y}\frac{\text{dy}}{\text{dx}} = \text{3ax}^{2}\therefore \frac{\text{dy}}{\text{dx}} = \frac{\text{3a}}{2} \frac{\text{x}^{2}}{\text{y}}$
Slope of tangent at (2, 3) $\frac{\text{dy}}{\text{dx}}\bigg]_{(2, 3)} =\frac{\text{3a}}{2}. \frac{4}{3} = \text{2a}$
Comparing with slope of tangent $\text{y = 4x - 5, we get, 2a = 4}\therefore$
Also (2, 3) lies on the curve $\therefore 9 = \text{8a + b, put a = 2, we get b = -7}$
View full question & answer→Question 1545 Marks
Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x - 3y + 2z - 5 = 0 and 2x - y + 3z - 1 = 0 and passing through (1, -2, 3).
AnswerThe equation of the plane passing through the line of intersection of the given planes is
$\text{x}-3\text{y}+2\text{z}-5+\lambda(2\text{x}-\text{y}+3\text{z}-1)=0\ ...(\text{i})$
This passing through (1, -2, 3). So,
$1+6+6-5+\lambda(2+2+9-1)$
$\Rightarrow8+12\lambda=0$
$\Rightarrow\lambda=\frac{-2}{3}$
Substituting this in (i) we get
$\text{x}-3\text{y}+2\text{z}-5-\frac{2}{3}(2\text{x}-\text{y}+3\text{z}-1)=0$
$\Rightarrow-\text{x}-7\text{y}-13=0$
$\Rightarrow\text{x}+7\text{y}+13=0$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+7\hat{\text{j}})+13=0,$ Which is the required vector equation of the plane.
View full question & answer→Question 1555 Marks
Find the equation of the plane through the point $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and passing throught the line of intersection of the planes $\vec{\text{r}}\cdot(\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=0$ and $\vec{\text{r}}\cdot(\hat{\text{j}}+2\hat{\text{k}})=0.$
AnswerThe equation of the plane passing through the line of intersection of the given planes is,
$\vec{\text{i}}(\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})+\lambda\big(\vec{\text{r}}(\hat{\text{j}}+2\hat{\text{k}})\big)=0$
$\vec{\text{r}}\cdot\Big[\hat{\text{i}}+(3+\lambda)\hat{\text{j}}+(-1+2\lambda)\hat{\text{k}}\Big]=0\ ...(\text{i})$
This passes through $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ So,
$(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})\Big[\hat{\text{i}}+(3+\lambda)\hat{\text{j}}+(-1+2\lambda)\hat{\text{k}}\Big]=0$
$\Rightarrow2+3+\lambda+1-2\lambda=0$
$\Rightarrow\lambda=6$
Substituting this in (i), we get
$\vec{\text{r}}\cdot\Big[\hat{\text{i}}+(3+6)\hat{\text{j}}+(-1+12)\hat{\text{k}}\Big]=0$
$\Rightarrow\vec{\text{r}}(\hat{\text{i}}+9\hat{\text{j}}+11\hat{\text{k}})=0$
View full question & answer→Question 1565 Marks
The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} - 5\hat{k} \text{ and } 2\hat{i} + 2\hat{j} + 3\hat{k}.$ Find the two unit vectors parallel its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Answer$\text{let } \text{d}_{1} \& \text{ d}_{2}$ be the two diagonal vectors:
$\therefore \overrightarrow{\text{d}}_{1} = 4\hat{\text{i}} - 2\hat{\text{j}} - 2\hat{\text{k}}, \overrightarrow{\text{d}}_{2} = -6\hat{\text{j}} - 8\hat{\text{k}}$
$\text{or} \overrightarrow{\text{d}}_{2} = 6\hat{\text{j}} + 8\hat{\text{k}}$
Unit vectors parallel to the diagonals are:
$\overrightarrow{\text{d}}_{1} = \frac{2}{\sqrt{6}}\hat{\text{i}} - \frac{1}{\sqrt{6}}\hat{\text{j}} - \frac{1}{\sqrt{6}}\hat{\text{k}}$
$\overrightarrow{\text{d}}_{2} = \frac{3}{5}\hat{\text{j}} - \frac{4}{5}\hat{\text{k}} \bigg(\overrightarrow{\text{d}}_{2} = \frac{3}{5}\hat{\text{j}} + \frac{4}{5}\hat{\text{k}}\bigg)$
$\overrightarrow{\text{d}}_{1}\times\overrightarrow{\text{d}}_{2} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 4 & -2 & -2 \\ 0 & -6 & -8 \end{vmatrix} = 4\hat{\text{i}} + 32\hat{\text{j}} - 24\hat{\text{k}} $
Area of parallelogram $= \frac{1}{2}|\overrightarrow{\text{d}}_{1}\times\overrightarrow{\text{d}}_{2}| = \sqrt{404} \text{ or } 2\sqrt{101} \text{sq. units}$
View full question & answer→Question 1575 Marks
If the lines $\text{x}=5,\frac{\text{y}}{3-\alpha}=\frac{\text{z}}{-2}$ and $\text{x}=\alpha,\frac{\text{y}}{-1}=\frac{\text{z}}{2-\alpha}$ are coplanar, find the values of $\alpha.$
AnswerThe lines $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ are coplanar if
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
The given lines $\frac{\text{x}-5}{0}=\frac{\text{y}}{3-\alpha}=\frac{\text{z}}{-2}$ and $\frac{\text{x}-\alpha}{0}=\frac{\text{y}}{-1}=\frac{\text{z}}{2-\alpha}$ are coplanar.
$\therefore\ \begin{vmatrix} \alpha-5&0-0&0-0\\0&3-\alpha&-2\\0&-1&2-\alpha\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix} \alpha-5&0-0&0-0\\0&3-\alpha&-2\\0&-1&2-\alpha\end{vmatrix}=0$
$\Rightarrow(\alpha-5)\Big[(3-\alpha)\times(2-\alpha)-2\Big]-0+0=0$
$\Rightarrow(\alpha-5)(\alpha-1)(\alpha-4)=0$
$\Rightarrow\alpha-1=0\text{ or }\alpha-4=0\text{ or }\alpha-5=0$
$\Rightarrow\alpha=1\text{ or }\alpha=4\text{ or }\alpha=5$
Thus, the values of $\alpha$ are 1, 4 and 5.
View full question & answer→Question 1585 Marks
$\text{Let } \vec{\text a} = \hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}, \vec{\text{b}} = \hat{\text{i}} \text{ and } \vec{\text{c}} = \text{c}_{1} \hat{\text{i}} + \text{c}_{2} \hat{\text{j}} + \text{c}_{3} \hat{\text{k}}, \text{then}$
- Let $c_1 = 1$ and $c_2 = 2$, find $c_3$ which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
- If $c_2 = –1$ and $c_3 = 1$, show that no value of $c_1$ can make $\vec{\text{a}}, \vec{\text{b}} \text{ and } \vec{\text{c}} \text{ coplanar}.$
Answer$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ \text{c}_{1} & \text{c}_{2} & \text{c}_{3} \end{vmatrix} = \text{c}_{2} - \text{c}_{3}$
- $\text{c}_{1} = 1, \text{ c}_{2} = 2$
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 2 - \text{c}_{3}$
$\therefore \vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ are coplanar} [\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 0 \Rightarrow \text{c}_{3} = 2$
- $\text{c}_{2} = -1, \text{c}_{3} = 1$
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \text{c}_{2} - \text{c}_{3} = -2 \neq 0$
$\Rightarrow \text{No value of }\text{c}_{1} \text{can make }\vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ coplanar}.$ View full question & answer→Question 1595 Marks
Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, -4, -5) and B(2, -3, 1) intersects the plane 2x + y + z = 7.
AnswerThe equation of a line joining the points A(3, -4, -5) and B(2, -3, 1) is
$\frac{\text{x}-3}{2-3}=\frac{\text{y}+4}{-3+4}=\frac{\text{z}+5}{1+5}=\text{r}$
$\Rightarrow\text{x}=3-\text{r},\text{ y}=-4+\text{r},\text{ z}=-5+6\text{r}$
Substituting this into the given plane equation we get
$2(3-\text{r})+(-4+\text{r})+(-5+6\text{r})=7$
$\Rightarrow\text{r}=2$
$\Rightarrow\text{x}=1,\text{ y}=-2,\text{ z}=7$
Distance of (1, -2, 7) from (3, 4, 4) is
$=\sqrt{(3-1)^2+(4+2)^2+(4-7)^2}$
$=\sqrt{4+36+9}$
$=\sqrt{49}$
$=7$
View full question & answer→Question 1605 Marks
Find the cartesian form of the equations of the following planes.
$\vec{\text{r}}=(1+\text{s}+\text{t})\hat{\text{i}}+(2-\text{s}+\text{t})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
AnswerThe given equation of the plane is,
$\vec{\text{r}}=(1+\text{s}+\text{t})\hat{\text{i}}+(2-\text{s}+\text{t})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+\text{s}(\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})+\text{t}(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-1&-2\\1&1&2\end{vmatrix}$
$=0\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}$
$=-4\hat{\text{j}}+2\hat{\text{k}}$
The vector equation of the plane in scalar product from is
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(-4\hat{\text{j}}+2\hat{\text{k}})=(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})(-4\hat{\text{j}}+2\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\Big[-2(2\hat{\text{j}}-\hat{\text{k}})\Big]=0-8+6$
$\Rightarrow\vec{\text{r}}\cdot\Big[-2(2\hat{\text{j}}-\hat{\text{k}})\Big]=-2$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{j}}-\hat{\text{k}})=1$
For cartesian form, let us substitute $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ here. Then, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{j}}-\hat{\text{k}})=1$
$\Rightarrow2\text{y}-\text{z}=1$
View full question & answer→Question 1615 Marks
Find the equation of the plane passing through the line of intersection of the planes $x + 2y +3z - 4 = 0$ and $2x + y - z + 5 = 0$ and perpendicular to the plane $5x + 3y - 6z + 8 = 0$.
AnswerThe equationof the planepassing the line of intersection of the given planes is,
$\text{x}+2\text{y}+3\text{z}-4+\lambda(2\text{x}+\text{y}-\text{z}+5)=0$
$(1+2\lambda)\text{x}+(2+\lambda)\text{y}+(3-\lambda)\text{z}-4+5\lambda=0\ ...(\text{i})$
The plane is perpendicular to 5x + 3y - 6z + 8 = 0. So,
$5(1+2\lambda)+3(2+\lambda)-6(3-\lambda)=0$ (Because $a_1a_2 + b_1b_2 + c_1c_2= 0$)
$\Rightarrow5+10\lambda+6+3\lambda-18+6\lambda=0$
$\Rightarrow19\lambda-7=0$
$\Rightarrow\lambda=\frac{7}{19}$
Substituting this in (i), we get
$\Big(1+2\Big(\frac{7}{19}\Big)\Big)\text{x}+\Big(2+\frac{7}{19}\Big)\text{y}+\Big(3-\frac{7}{19}\Big)\text{z}-4+5\Big(\frac{7}{19}\Big)=0$
$\Rightarrow33\text{x}+45\text{y}+50\text{z}-41=0$
View full question & answer→Question 1625 Marks
Find the equation of a plane passing through the points $(0, 0, 0)$ and $(3, -1,2)$ and parallel to the line $\frac{\text{x}-4}{1}=\frac{\text{y}+3}{-4}=\frac{\text{z}+1}{7}$
AnswerThe equation of the plane through $(0, 0, 0) $ is
$a(x - 0) + b(y - 0) + c(z - 0) = 0$
$ax + by + cz = 0 ...(i)$
This plane passes through (3, -1, 2) So,
$3a - b + 2c = 0 ....(ii)$
Again plane (i) is parallel to the given line.
It means that the normal to plane (i) is perpendicular to the line.
$\Rightarrow a(1) + b(-4) + c(7) = 0 ....(iii)$ (Because $a_1a_2 + b_1b_2 + c_1c_2= 0)$
Solving (i), (ii) and (iii) we get
$\begin{vmatrix}\text{x}&\text{y}&\text{z}\\3&-1&2\\1&-4&7\end{vmatrix}=0$
$\Rightarrow x - 19y - 11z = 0$
View full question & answer→Question 1635 Marks
Find the angle between the vectors with direction ratios proportional to 1, -2, 1 and 4, 3, 2.
AnswerLet $\vec{\text{a}}$ be a vector with direction ratios 1, -2, 1.
$\Rightarrow\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
Let $\vec{\text{b}}$ be a vector with direction ratios 4, 3, 2.
$\Rightarrow\vec{\text{b}}=4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Let $\theta$ be the angle between the given vectors.
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\big|\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big|\big|4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\big|}$
$=\frac{4-6+2}{\sqrt{1+4+1}\sqrt{16+9+4}}$
$=\frac{0}{\sqrt{6}\sqrt{29}}$
$=0$
$\therefore\theta=\frac{\pi}{2}$
Thus, the angle between the given vectors measures $\frac{\pi}{2}$.
View full question & answer→Question 1645 Marks
Find the shortest distance between the following pairs of parallel lines whose equations are:$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
AnswerThe vector equation of the given lines are
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)-\mu\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
These two lines pass through the points having position vectors $\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{a}}_2=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$ and are parallel to the vector $\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
Now,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}-3\hat{\text{j}}-4\hat{\text{k}}\big)\times\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&-4\\1&-1&1 \end{vmatrix}$
$=-7\hat{\text{i}}-5\hat{\text{j}}+2\hat{\text{k}}$
$\Rightarrow\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|=\sqrt{(-7)^2+(-5)^2+2^2}$
$=\sqrt{49+25+4}$
$=\sqrt{78}$
The shortest distance between the two lines is given by
$\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|}{\big|\vec{\text{b}}\big|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$
View full question & answer→Question 1655 Marks
Find the value of $\lambda,$ so that the lines $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$ are at right angles. Also, find whether the lines are intersecting or not.
Answer Given lines are $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$
Converting them into standard form,
we have $=\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{\Big(\frac{\lambda}{7}\Big)}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\Big(\frac{-3\lambda}{7}\Big)}=\frac{\text{y}-5}{1}=\frac{\text{z}-6}{-5}$
View full question & answer→Question 1665 Marks
Find the value of $\lambda,$ so that the lines $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$ are at right angles. Also, find whether the lines are intersecting or not.
AnswerGiven lines are $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$
Converting them into standard form,
we have $=\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{\Big(\frac{\lambda}{7}\Big)}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\Big(\frac{-3\lambda}{7}\Big)}=\frac{\text{y}-5}{1}=\frac{\text{z}-6}{-5}$
View full question & answer→Question 1675 Marks
Find the equation of the plane through the line of intersection of the planes $x + 2y + 3z + 4 = 0$ and $x - y + z + 3 = 0$ and passing through the origin.
AnswerWe know that, equation of a plane passing through the line of intersection of planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by,
$(\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1)+\lambda(\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2)=0$
So, equation of plane passing through the line of intersection of planes $x + 2y + 3z + 4 = 0$ and $x - y + z + 3 = 0$ is
$(\text{x}+2\text{y}+3\text{z}+4)+\lambda(\text{x}-\text{y}+\text{z}+3)=0$
$\text{x}(1+\lambda)+\text{y}(2-\lambda)+\text{z}(3+\lambda)+4+3\lambda=0\ ...(\text{i})$
Equation (i) is passing through origin, so
$(0)(1+\lambda)+(0)(2-\lambda)+(0)(3+\lambda)+4+3(\lambda)=0$
$0+0+0+4+3\lambda=0$
$3\lambda=-4$
$\lambda=-\frac{4}{3}$
Put the value of $\lambda$ in equation (i),
$\text{x}(1+\lambda)+\text{y}(2-\lambda)+\text{z}(3+\lambda)+4+3\lambda=0$
$\text{x}\Big(1-\frac{4}{3}\Big)+\text{y}\Big(2+\frac{4}{3}\Big)+\text{z}\Big(3-\frac{4}{3}\Big)+4-\frac{12}{3}=0$
$\text{x}\Big(\frac{3-4}{3}\Big)+\text{y}\Big(\frac{6+4}{3}\Big)+\text{z}\Big(\frac{9-4}{3}\Big)+4-4=0$
$-\frac{\text{x}}{3}+\frac{10\text{y}}{3}+\frac{5\text{z}}{3}=0$
Multiplying by 3, we get
$-x + 10y + 5z = 0$
$x - 10y - 5z = 0$
The equation of required plane is,
$x - 10y - 5z = 0$
View full question & answer→Question 1685 Marks
Find the shortest distance between the lines:
$\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}\ \text{and}\ \frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}.$
AnswerEquation of one line is $\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}$
Comparing this equation with $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1},$ we have
$x_1 = -1, y_1 = -1, z_1 = -1,$
$a_1 = 7, b_1 = -6, c_1 = 1$
Comparing this equation with $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2},$ we have
$x_2 = 3, y_2 = 5, z_2 = 7,$
$a_2 = 1, b_2 = -2, c_2 = 1$
$\therefore\ \ \begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}$
$=\begin{vmatrix}3+1&5+1&7+1\\7&-6&1\\1&-2&1\end{vmatrix}$
Expanding first row = 4(-6 + 2) - 6(7 - 1) + 8(-14 + 6) = -16 - 36 - 64 = -116
And $\sqrt{(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)+(\text{b}_1\text{c}_2-\text{b}_2\text{c}_1)+(\text{c}_1\text{a}_2-\text{c}_2\text{a}_1)}$
$=\sqrt{(-14+6)^2+(-6+2)^2+(1-7)^2}$
$=\sqrt{64+16+36}=\sqrt{116}$
$\therefore$ Length of shortest distance
$=\frac{\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}}{\sqrt{(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)+(\text{b}_1\text{c}_2-\text{b}_2\text{c}_1)+(\text{c}_1\text{a}_2-\text{c}_2\text{a}_1)}}$
$=\frac{-116}{\sqrt{116}}=-\sqrt{116}=\sqrt{116}\ (\text{numerically})$
$=\sqrt{4\times29}=2\sqrt{29}$ View full question & answer→Question 1695 Marks
Find the equations of the planes parallel to the plane x - 2y + 2z - 3 = 0 and which are at a unit distance from the point (1, 1, 1).
AnswerThe equation of the plane parallel to the given plane is
x - 2y + 2z + k = 0 ...(i)
It is given the plane (i) is at a distance of 1 unit from (1, 1, 1)
$\Rightarrow\frac{|1-2+2+\text{k}|}{\sqrt{1^2+(-2)^2+2^2}}=1$
$\Rightarrow\frac{|1+\text{k}|}{3}=1$
⇒ |1 + k| = 3
⇒ 1 + k = 3, 1 + k = -3
Substituting these two values one by one in (i) we get
x - 2y + 2z + 2 = 0 and x - 2y + 2z - 4 = 0, which are the equations of the required planes.
View full question & answer→Question 1705 Marks
Find the cartesian form of the equations of the following planes.
$\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}})+\text{s}(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})+\text{t}(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})$
Answer$\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}})+\text{s}(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})+\text{t}(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})$We know that the equation $\vec{\text{r}}=\vec{\text{a}}+\text{s}\vec{\text{b}}+\text{t}\vec{\text{c}}$ represents a plane passing through a point whose position vector is $\vec{\text{a}}$ and parallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}.$
Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&2\\1&2&1\end{vmatrix}$
$=-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
The vector equation of the plane in scalar product form is,
$\vec{\text{i}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})=(\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}})\cdot(-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\Big[-3(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\Big]=-3-3+0 $
$\Rightarrow\vec{\text{r}}\cdot\Big[-3(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\Big]=-6$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=2$
For cartesian form, let us substitute $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ here. Then we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=2$
$\text{x}-\text{y}+\text{z}=2$
View full question & answer→Question 1715 Marks
Find the equation of the plane that contains the point (1, -1, 2) and is perpendicular to each of the planes 2x + 3y - 2z = 5 and x + 2y - 3z = 8.
AnswerThe equation of any plane passing through (1, -1, 2) is,
a(x - 1) + b(y + 1) + c(z - 2) = 0 ....(i)
It is given that (i) is perpendicular to the plane 2x + 3y - 3z = 5. So,
2a + 3b - 3c = 0 ....(ii)
It is given that (i) is perpendicular to the plane x + 2y - 3z = 8. So,
a + 2b - 3c = 0 ....(iii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}-1&\text{y}+1&\text{z}-2\\2&3&-2\\1&2&-3\end{vmatrix}=0$
⇒ -5(x - 1) + 4(y + 1) + 1(z - 2) = 0
⇒ 5x - 4y - z = 7
View full question & answer→Question 1725 Marks
Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines$\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{3}$ and$\frac{\text{x}}{-3}=\frac{\text{y}}{2}=\frac{\text{z}}{5}.$
AnswerLet theD.R’s of the required line be a,b , c $\therefore\ $a + 2b + 3c = 0 and –3a + 2b + 5c = 0 $\Rightarrow\ \frac{\text{a}}{4}=\frac{\text{b}}{-14}=\frac{\text{c}}{8}\ \therefore\ \text{DRS are} \ 2,-7,4$$\therefore\ $ Equations of line are $\frac{\text{x}-2}{2}=\frac{\text{y}-1}{-7}=\frac{\text{z-3}}{4}$
which, in vector form is,$\overrightarrow{\text{r}}=(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})+\lambda(2\hat{\text{i}}-7\hat{\text{j}}+4\hat{\text{k}})$
View full question & answer→Question 1735 Marks
Find the cartesian and vector equations of a line which passes through the point $(1, 2, 3)$ and is parallel to the line $\frac{-\text{x}-2}{1}=\frac{\text{y}+3}{7}=\frac{2\text{z}-6}{3}.$
Answerwe know that, equation of a line passing through a point $(x_1, y_1, z_1)$ and having direction ratios proportional to a, b, c is
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}\dots(1)$
Here, $(x_1, y_1, z_1) = (1, 2, 3)$ and
Given line $\frac{-\text{x}-2}{1}=\frac{\text{y}+3}{7}=\frac{2\text{z}-6}{3}$
$\Rightarrow\frac{\text{x}+2}{-1}=\frac{\text{y}+3}{7}=\frac{\text{z}-6}{\frac{3}{2}}$
It parallel to the required line, so
$\text{a}=\mu,\text{b}7\mu,\text{c}=\frac{3}{2}\mu$
So, equation of required line using equation (1) is,
$\frac{\text{x}-1}{-\mu}=\frac{\text{y}-2}{7\mu}=\frac{\text{z}-3}{\frac{3}{2}\mu}$
$\Rightarrow\frac{\text{x}-1}{-1}=\frac{\text{y}-2}{7}=\frac{\text{z}-3}{\frac{3}{2}}$
View full question & answer→Question 1745 Marks
The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} - 5\hat{k} \text{ and } 2\hat{i} + 2\hat{j} + 3\hat{k}.$ Find the two unit vectors parallel its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Answer$\text{let } \text{d}_{1} \& \text{ d}_{2}$ be the two diagonal vectors:
$\therefore \overrightarrow{\text{d}}_{1} = 4\hat{\text{i}} - 2\hat{\text{j}} - 2\hat{\text{k}}, \overrightarrow{\text{d}}_{2} = -6\hat{\text{j}} - 8\hat{\text{k}}$
$\text{or} \overrightarrow{\text{d}}_{2} = 6\hat{\text{j}} + 8\hat{\text{k}}$
Unit vectors parallel to the diagonals are:
$\overrightarrow{\text{d}}_{1} = \frac{2}{\sqrt{6}}\hat{\text{i}} - \frac{1}{\sqrt{6}}\hat{\text{j}} - \frac{1}{\sqrt{6}}\hat{\text{k}}$
$\overrightarrow{\text{d}}_{2} = \frac{3}{5}\hat{\text{j}} - \frac{4}{5}\hat{\text{k}} \bigg(\overrightarrow{\text{d}}_{2} = \frac{3}{5}\hat{\text{j}} + \frac{4}{5}\hat{\text{k}}\bigg)$
$\overrightarrow{\text{d}}_{1}\times\overrightarrow{\text{d}}_{2} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 4 & -2 & -2 \\ 0 & -6 & -8 \end{vmatrix} = 4\hat{\text{i}} + 32\hat{\text{j}} - 24\hat{\text{k}} $
Area of parallelogram $= \frac{1}{2}|\overrightarrow{\text{d}}_{1}\times\overrightarrow{\text{d}}_{2}| = \sqrt{404} \text{ or } 2\sqrt{101} \text{sq. units}$
View full question & answer→Question 1755 Marks
Find the cartesian equation of a line passing through (1, -1, 2) and parallel to the line whose equation are $\frac{\text{x}-3}{1}=\frac{\text{y}-1}{2}=\frac{\text{z}+1}{-2}.$ Also, reduce the equation obtained in vector form.
View full question & answer→Question 1765 Marks
Find the intercepts made on the coordinate axes by the plane 2x + y - 2z = 3 and also find the direction cosines of the normal to the plane.
AnswerHere, given equation of plane is,
2x + y - 2z = 3
Dividing by 3 on both the sides,
$\frac{2\text{x}}{3}+\frac{\text{y}}{3}-\frac{2\text{z}}{3}=\frac{3}{3}$
$\frac{\text{x}}{\frac{3}{2}}+\frac{\text{y}}{3}+\frac{\text{z}}{-\frac{3}{2}}=1\ ...(\text{i})$
We know that, if a, b, c are the intercepts by a plane on the coordinate axes,
new equation of the plane is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(\text{ii})$
Comparing the equation (i) and (ii),
$\text{a}=\frac{3}{2},\text{b}=3,\text{c}=-\frac{3}{2}$
Again, given equation of plane is,
2x + y - 2z = 3
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})=3$
$\vec{\text{r}}(2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})=3$
So, vector normal to the plane is given by
$\vec{\text{n}}=2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{(2)^2+(1)^2+(-2)^2}$
$=\sqrt{4+1+4}$
$=\sqrt{9}$
$|\vec{\text{n}}|=3$
Direction vector of $\vec{\text{n}}=2,1,-2$
Direction vector $\vec{\text{n}}=\frac{2}{|\vec{\text{n}}|},\frac{1}{|\vec{\text{n}}|},\frac{-2}{|\vec{\text{n}}|}$
$=\frac{2}{3},\frac{1}{3},-\frac{2}{3}$
So,
Intercepts by the plane on coordinaye axes are $=\frac{3}{2},3,-\frac{3}{2}$
Direction cosine of normal to the plane are $=\frac{2}{3},\frac{1}{3},-\frac{2}{3}$
View full question & answer→Question 1775 Marks
Find the equation of a plane which passes through the point (3, 2, 0) and contains the line $\frac{\text{x}-3}{1}=\frac{\text{y}-6}{5}=\frac{\text{z}-4}{4}$
AnswerRequired equation of plane is passing through the point (3, 2, 0)
$\therefore$ a(x - 3) + b(y - 2) + c(z - 0) = 0
⇒ a(x - 3) + b(y - 2) + cz = 0 ....(i)
Required equation of plane also contains the line $\frac{\text{x}-3}{1}=\frac{\text{y}-6}{5}=\frac{\text{z}-4}{4},$ so it passes through the point (3, 2, 0)
⇒ a(3 - 3) + b(6 - 2) + c(4) = 0
⇒ 4b + 4c = 0 ...(ii)
Also plane will be parallel to,
a(1) + b(5) + c(4) = 0
a + 5b + 4c = 0 ....(iii)
Solving (ii) and (iii) by cross multiplication,
$\frac{\text{a}}{16-20}=\frac{\text{b}}{4-0}=\frac{\text{c}}{0-4}=\lambda(\text{say})$
$-\frac{\text{a}}{4}=\frac{\text{b}}{4}=-\frac{\text{c}}{4}=\lambda(\text{say})$
$\Rightarrow\text{a}=-\lambda,\text{ b}=\lambda,\text{ c}=-\lambda$
Put $\text{a}=-\lambda,\text{ b}=\lambda,\text{ c}=-\lambda$ in equation (i) we get
$(-\lambda)(\text{x}-3)+(\lambda)(\text{y}-2)+(-\lambda)\text{z}=0$
$\Rightarrow-\text{x}+3+\text{y}-2-\text{z}=0$
$\Rightarrow\text{x}-\text{y}+\text{z}-1=0$
View full question & answer→Question 1785 Marks
Find the coordinates of the foot of the perependicular drawn from the origin to the plane 2x - 3y + 4z - 6 = 0.
AnswerLet M be the foot of the perpendicular of the origin P(0, 0, 0) in the plane 2x - 3y + 4z - 6 =0.
Then, PM is normal to the plane. So, the direction rations of PM are proportional to 2, -3, 4.
Since PM passes through P(0, 0, 0) and has direction ratios proportional to 2, -3, 4 the equation ot PQ is
$\frac{\text{x}-0}{2}=\frac{\text{y}-0}{-3}=\frac{\text{z}-0}{4}=\text{r (say)}$
Let the coordinted of M be (2r, -3r, 4r)
Since M lies in the plane 2x - 3y + 4z - 6 = 0,
2(2r) - 3(-3r) + 4(4r) - 6 = 0
⇒ 4r + 9r + 16r - 6 = 0
⇒ 29r - 6 = 0
$\Rightarrow\ \text{r}=\frac{6}{29}$
Substituting the value of r in the coordinated of Ml we get
$\text{M}=(2\text{r},-3\text{r},4\text{r})=\bigg(2\Big(\frac{6}{29}\Big),-3\Big(\frac{6}{29}\Big),4\Big(\frac{6}{29}\Big)\bigg)$
$=\Big(\frac{12}{29},\frac{-18}{29},\frac{24}{29}\Big)$
View full question & answer→Question 1795 Marks
Show that the vectors $\overrightarrow{\text{a}},\overrightarrow{\text{b}} \text{and}{\overrightarrow{\text{c}}}$ are coplanar $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}}+\overrightarrow{\text{c}}\text{and} \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar.
AnswerGiven that $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar
$\therefore[\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}] = 0$
$\text{i.e} \overrightarrow{\text{(a}} +\overrightarrow{\text{b})}, \big\{\overrightarrow{\text{(b}} + \overrightarrow{\text{c})} \times\overrightarrow{\text{(c}} + \overrightarrow{\text{a})}\big\} = 0$
$\overrightarrow{\text{(a}} +\overrightarrow{\text{b})}. \big\{\overrightarrow{\text{(b}} \times \overrightarrow{\text{c}}+ \overrightarrow{\text{b}} \times\overrightarrow{\text{a}}\times\overrightarrow{\text{c}} \times \overrightarrow{\text{a})}\big\} = 0$
$\Rightarrow\overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{a}}(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})}+\overrightarrow{\text{b}}. (\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})} = 0$
$\Rightarrow 2 [\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}] = \text{0 or} [\overrightarrow{\text{a}},\overrightarrow{\text{b}}, \overrightarrow{\text{c] }} = 0$
$\Rightarrow\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}\text{are coplanar}.$
View full question & answer→Question 1805 Marks
Find the angle between the lines whose direction cosines are given by the equations $l + m + n = 0, l^2 + m^2- n^2 = 0.$
AnswerEliminating n from both the equations, we have $l^2 + m^2 - (l - m)^2 = 0$
$\Rightarrow l^2 + m^2 - l^2 - m^2 + 2ml = 0$
$\Rightarrow 2lm = 0$
$\Rightarrow lm = 0$
$\Rightarrow (-m - n)m = 0$
$[\because\text{l}=-\text{m}-\text{n}]$
$\Rightarrow m = -n$
$\Rightarrow m = 10$
$\Rightarrow l = 0, l = -n$
Thus, Dr’s two lines are proportional to 0, -n, n and -n, 0, -1 i.e., 0, -1, 1 and -1, 0, 1.So, the vector parallel to these given lines are $\vec{\text{a}}=-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{k}}$
Now, $\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{|\vec{\text{a}}||\vec{\text{b}}|}\Rightarrow\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}\Big[\because\cos\frac{\pi}{3}=\frac{1}{2}\Big]$
View full question & answer→Question 1815 Marks
In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
$2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$
AnswerThe direction ratios of normal to the plane, $L_1: a_1x + b_1y + c_1z = 0,$
are $a_1, b_1, c_1$_ and $L_2: a_2x + b_2y + c_2z = 0$ are $a_2, b_2, c_2$
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between L_1 and L_2 is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}.\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$
The equations of the given planes are $2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$
Here, $a_1 = 2, b_1 = -2, c_1 = 4$ and $a_2 = 3, b_2 = -3, c_2 = 6$
$a_1a_2 +b_1b_2 + c_1c_2 = 2 \times 3 + (-2) \times (-3) + 4 \times 6 = 6 + 6 + 24$
$=36\neq0$
Thus, the given planes are not perpendicular to each other.
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{3},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-2}{-3}=\frac{2}{3}\text{ and } \frac{\text{c}_1}{\text{c}_2}=\frac{4}{6}=\frac{2}{3}$
$\therefore\ \ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Thus, the given planes are parallel to each other.
View full question & answer→Question 1825 Marks
Find the equation of the plane mid-parallel to the planes $2x - 2y + z + 3 = 0$ and $2x - 2y + z + 9 = 0$
AnswerWe know that the distance between two planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is $\frac{\big|\text{d}_2-\text{d}_1\big|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
The equation of plane thatb is mid-parallel to the planes
$2x - 2y + z + 3 = 0 ...(i)$
$2x - 2y + z + 9 = 0 ....(ii)$
is of the form $2x - 2y + z + k = 0 ...(iii)$
It meance that the distance between (i) and (iii) = distance between (i) and (ii)
$\Rightarrow\frac{|\text{k}-3|}{\sqrt{4+4+1}}=\frac{|\text{k}-9|}{\sqrt{4+4+1}}$
$\Rightarrow |k - 3| = |k - 9|$
$\Rightarrow k - 3 = k - 9 or k - 3 = -(k - 9)$
$\Rightarrow 3 = 9$ (false);$ k - 3 = -k + 9$
$\Rightarrow 2k = 12$
$\Rightarrow k = 6$
Substituting this in (iii) we get $2x - 2y + z + 6 = 0$, which is the required equation of the plane.
View full question & answer→Question 1835 Marks
Find the vector equation of the line through the origin which is perpendicular to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=3$
AnswerRequired line is perpendicular to plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=3,$ so line is parallel to the normal vector $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ of plane.
And it is passing through point $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
We know that equation of a passing through $\vec{\text{a}}$ and parallel to vector $\vec{\text{b}}$ is,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\ ...(\text{i})$
Here, $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and $\vec{\text{b}}=\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
So, $\vec{\text{r}}=(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
Hence, equation required line is
$\vec{\text{r}}=\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
View full question & answer→Question 1845 Marks
Find the angle between line $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}+1}{1}$ and the plane $2x + y - z = 4$.
AnswerWe know that the angle $\theta$ between the line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and plane $a_2x + b_2y + c_2z + d_2= 0$ is given by
$\sin\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}\ ...(\text{i})$
Given, equation of line is
$\frac{\text{x}-1}{1}=\frac{\text{y}-2}{-1}=\frac{\text{z}+1}{1}$
So, $a_1 = 1, b_1 = -1, c_1 = 1$
Given equation of plane is 2x + y - z - 4 = 0
So, $a_2 = 2, b_2 = 1, c_2 = -1$
Put these value in equation (i),
$\sin\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}^2_1+\text{b}^2_1+\text{c}^2_1}\sqrt{\text{a}^2_2+\text{b}^2_2+\text{c}^2_2}}$
$\sin\theta=\frac{(1)(2)+(-1)(1)+(1)(-1)}{\sqrt{(1)^2+(-1)^2+(1)^2}\sqrt{(2)^2+(1)^2+(-1)^2}}$
$\sin\theta=\frac{2-1-1}{\sqrt{1+1+1}\sqrt{4+1+1}}$
$\sin\theta=\frac{0}{\sqrt{3}\sqrt{6}}$
$\sin\theta=0$
$\theta=0^{\circ}$
angle between plane and line $=0^{\circ}$
View full question & answer→Question 1855 Marks
Find the angle between the lines whose direction cosines are given by the equations:
$2l + 2m - n = 0, mn + ln + lm = 0$
AnswerThe given equation are,
$2l + 2m - n = 0 .....(1)$
$mn + ln + lm = 0 .....(2)$
From (1), We get n = 2l + 2m.
Putting $n = 2l + 2m$ in $(2)$, We get
$m(2l + 2m) + l(2l + 2m) + lm = 0$
$2lm + 2m^2 + 2l^2 + 2ml + lm = 0$
$2ml^2 + 5lm + 2l^2 = 0$
$2m^2 + 4lm + lm + 2l^2 = 0$
$(2m + l) (m + 2l) = 0$
$\Rightarrow\text{m}=-\frac{1}{2}$ or $\text{m}=-2\text{l}$
By putting $\text{m}=-\frac{\text{l}}{2}$ in (1) we get n = l
By putting m = 2l in (i) we get n = -2l
So, vector parallel to these lines are
$\vec{\text{a}}=\hat{\text{i}}-\frac{1}{2}\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$ respectively.
If $\theta$ is the angle between the lines, then $\theta$ is also the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$
then,
$\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{1+1-2}{\sqrt{1+\frac{1}{4}+1}\sqrt{1+4+9}}=0$
$\theta=\cos^{-1}(0)=\frac{\pi}{2}$.
View full question & answer→Question 1865 Marks
Find the foot of the perpendicular from (0, 2, 7) on the line $\frac{\text{x}+2}{-1}=\frac{\text{y}-1}{3}=\frac{\text{z}-3}{-2}.$
AnswerLet L be the foot of the perpendicular drawn from the point P (0, 2, 7) to the given line. The coordoinates of a general point on the line $\frac{\text{x}+2}{-1}=\frac{\text{y}-1}{3}=\frac{\text{z}-3}{-2}$ are given by $\frac{\text{x}+2}{-1}=\frac{\text{y}-1}{3}=\frac{\text{z}-3}{-2}=\lambda$ $\Rightarrow\text{x}=-\lambda-2$ $\text{y}=3\lambda+1$ $\text{z}=-2\lambda+3$ Let the coordinates of L be $-\lambda-2,3\lambda+1,-2\lambda+3.$
The direction ratios of PL are proportional to $-\lambda-2-0,3\lambda+1-2,-2\lambda+3-7,$ i,e. $-\lambda-2,3\lambda-1,-2\lambda-4.$ The direction ratios of the given line are proportionl to -1, 3, -2, but PL is perpendicular to the given line. $\therefore-1-\lambda-2+33\lambda-1-2\lambda-4=0\Rightarrow\lambda=-12$ Substituting $\lambda=-12$ in $-\lambda-2,3\lambda+1.-2\lambda+3,$ we get the coordinates of L as -32, -12, 4. View full question & answer→Question 1875 Marks
The direction ratios of the perpendicular from the origin to a plane are 12, -3, 4 and the length of the perpendicular is 5. Find the equation of the plane.
AnswerGiven, direction ratios of perpendicular from origin to a plane is 12, -3, 4
So,
Normal vector $=12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{n}}=12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{(12)^2+(-3)^2+(4)^2}$
$=\sqrt{144+9+16}$
$=\sqrt{169}$
$|\vec{\text{n}}|=13$
Normal unit vector $\hat{\text{n}}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}$
$=\frac{1}{13}(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})$
Given that, perpendicular distance of plane from origin is 5 unit.
$\Rightarrow\text{d}=5\text{ unit}$
We know that, equation of a plane at a distance d from origin and normal unit vector $\hat{\text{n}}$ is,
$\vec{\text{r}}\cdot\hat{\text{n}}={\text{d}}$
So, vector equation of required plane is,
$\vec{\text{r}}\cdot\Big(\frac{12}{13}\hat{\text{i}}-\frac{3}{13}\hat{\text{j}}+\frac{4}{13}\hat{\text{k}}\Big)=5$
Put $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\Big(\frac{12}{13}\hat{\text{i}}-\frac{3}{13}\hat{\text{j}}+\frac{4}{13}\hat{\text{k}}\Big)=5$
$(\text{x})\Big(\frac{12}{13}\Big)+(\text{y})\Big(-\frac{3}{13}\Big)+(\text{z})\Big(\frac{4}{13}\Big)=5$
$\frac{12}{13}\text{x}-\frac{3}{13}\text{y}+\frac{4}{13}\text{z}=5$
View full question & answer→Question 1885 Marks
Find the angle between the pairs of lines with direction ratios proportional to
$2, 2, -2$ and $4, 1, 8$
AnswerWe know that, angle $(\theta)$ between lines
$\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$
is given by,
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{{\text{a}_1}^2+{{\text{b}_1}^2}+{{\text{c}_1}^2}}\sqrt{{\text{a}_2}^2+{{\text{b}_2}^2}+{{\text{c}_2}^2}}}\dots(1)$
Here, $a_1= 2, b_1= 2, c_1= 1$
$a_2= 4, b_2= 1, c_2= 8$
Let $\theta$ be required angle, so using equation (1),
$\cos\theta=\frac{(2)(4)+(2)(1)+(1)(8)}{\sqrt{(2)^2+(2)^2+(1)^2}\sqrt{(4)^2+(1)^2+(8)^2}}$
$=\frac{8+2+8}{3.9}$
$=\frac{18}{27}$
$\cos\theta=\frac{2}{3}$
$\theta=\cos^{-1}\Big(\frac{2}{3}\Big)$
View full question & answer→Question 1895 Marks
Find a vector of magnitude 26 units normal to the plane 12x - 3y + 4z = 1
AnswerGiven, equation of plane is,
12x - 3y + 4z = 1
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(12\hat{\text{i}}-3\hat{\text{j}}-4\hat{\text{k}})=1$
$\vec{\text{r}}\cdot\vec{\text{n}}=1$
So, normal to the plane is
$\vec{\text{n}}=12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{(12)^2+(-3)^2+(4)^2}$
$=\sqrt{144+9+16}$
$=\sqrt{144+25}$
$=\sqrt{169}$
$=13$
$\text{unit vector}\hat{\text{ n}}=\frac{12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}}{13}$
$=\frac{12\hat{\text{i}}}{13}-\frac{3}{13}\hat{\text{j}}+4\hat{\text{k}}$
A vector to the plane with magnitude
$26=26\hat{\text{n}}$
$=26\Big(\frac{12\hat{\text{i}}}{13}-3\hat{\text{j}}+4\hat{\text{k}}\Big)$
Required vector $=24\hat{\text{i}}-6\hat{\text{j}}+8\hat{\text{k}}$
View full question & answer→Question 1905 Marks
Find the length and the foot of the perpendicular drawn from the point $(2, –1, 5)$ to the line $\frac{\text{x - 11}}{10}=\frac{\text{y + 2}}{-4}=\frac{\text{z + 8}}{-11}.$
AnswerAny point $P$ on the line is given by
$x=10 \lambda+11, y=-4 \lambda-2, z=-11 \lambda-8$
The given point is $Q (2,-1,5)$
Direction Ratio's of PQ are $10 \lambda+9,-4 \lambda-1,-11 \lambda-13$
$PQ \perp$ to the given line
$\therefore 10(10 \lambda+9)-4(-4 \lambda-1)-11(-11 \lambda-13)=0$
$100 \lambda+90+16 \lambda+4+121 \lambda+143=0$
$237 \lambda+237=0 \Rightarrow \lambda=-1$
$\therefore$ The point P is $(11-10,4-2,11-8)$
OR $(1, 2, 3)$
$\therefore$ $PQ^2 = (2 - 1)^2 + (-1 - 2)^2 + (5 - 3)^2$
$= 1 + 9 + 4 = 14$
$\Rightarrow$ PQ = $\sqrt{14}$.
View full question & answer→Question 1915 Marks
ABCD is aparallelogram. the position vectora of the points A, B and C are respectively, $4\hat{\text{i}}+5\hat{\text{j}}-10\hat{\text{k}},2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}.$ Find the vector equation of the line BD. Also, reduce it to cartesian form.
AnswerWe know that the position vector of the mid-point of $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\vec{\text{a}}+\vec{\text{b}}}{2}.$
Let the position of mid-point of A and C= position vector of mid-point of B and D
$\therefore\frac{\big(4\hat{\text{i}}+5\hat{\text{j}}-10\hat{\text{k}}\big)+\big(-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2}=\frac{\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)}{2}$
$\Rightarrow\frac{3}{2}\hat{\text{i}}+\frac{7}{2}\hat{\text{j}}-\frac{9}2{}\hat{\text{k}}=\Big(\frac{\text{x}+2}{2}\Big)\hat{\text{i}}+\Big(\frac{-3+\text{y}}{2}\Big)\hat{\text{j}}+\Big(\frac{4+\text{z}}{2}\Big)\hat{\text{k}}$
Comparing the coeffient of $\hat{\text{i}},\hat{\text{j}}$ and $\hat{\text{k}},$ we get
$\frac{\text{x}+2}{2}=\frac{3}{2}$
$\Rightarrow\text{x}=1$
$\frac{-3+\text{y}}{2}=\frac{7}{2}$
$\Rightarrow\text{y}=10$
$\frac{4+\text{z}}{2}=-\frac{9}{2}$
$\Rightarrow\text{z}=-13$
Position vector of point $\text{D}=\hat{\text{i}}+10\hat{\text{j}}-13\hat{\text{k}}$
View full question & answer→Question 1925 Marks
Show that the lines $\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{4}$ and $\frac{\text{x}-4}{5}=\frac{\text{y}-1}{2}=\text{z}$ intersect. Also, find their point of intersection.
AnswerWe have lines
$\text{L}_1:\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{4}=\lambda$
and $\text{L}_2:\frac{\text{x}-4}{5}=\frac{\text{y}-1}{2}=\text{z}=\mu$
Any point on the line $L_1$ is $(2\lambda+1,3\lambda+2,4\lambda+3)$
Any point on the line $L_2$ is $(5\mu+4,2\mu+1,\mu)$
$(2\lambda+1,3\lambda+2,4\lambda+3)=(5\mu+4,2\mu+1,\mu)$
$\Rightarrow2\lambda+1=5\mu+4,3\lambda+2=2\mu+1$ and $4\lambda+3=\mu$
Solving first two equations we get $\lambda=-1,\mu=-1$
These values of $\lambda=-1,\mu=-1$ also satisfy the third equation.
Thus lines intersect.
Also the point of intersection is $(-1, -1, -1)$.
View full question & answer→Question 1935 Marks
Find the equation of the plane through (3, 4, -1) which is parallel to the plane $\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})+2=0$
AnswerLet the equation of a plane parallel to the given plane be
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})=\text{k}\ ...(\text{i})$
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})=\text{k}$
This passes through (3, 4, -1). so,
$(3\hat{\text{i}}+4\hat{\text{j}}-\hat{\text{k}})(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})=\text{k}$
$\text{k}=6-12-5=-11$
Substituting this in (i), we get
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})=-11$
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})+11=0,$ which is the equation of the required plane.
View full question & answer→Question 1945 Marks
If the line drawn from (4, -1, 2) meets a plane at right at the point (-10, 5, 4) find the equation of the plane.
AnswerThe normal is passing through the point A(4, -1, 2) and B(-10, 5, 4) So,
$\overrightarrow{\text{n}}=\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{OA}}$
$=(-10\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})-(4\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})$
$=-14\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}}$
Since the plane passes through (-10, 5, 4), $\vec{\text{a}}=-10\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is
$\vec{\text{r}}\cdot(-14\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}})=(-10\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})(-14\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(-14\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}})=140+30+8$
$\Rightarrow\vec{\text{r}}\cdot\big(-2(7\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}})\big)=178$
$\Rightarrow\vec{\text{r}}\cdot(7\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}})=-89$
Substituting $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(7\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}})=-89$
$\Rightarrow7\text{x}-3\text{y}-\text{z}=-89$
$\Rightarrow7\text{x}-3\text{y}-\text{z}+89=0$
View full question & answer→Question 1955 Marks
Find the shortest distance between the lines
$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$ $\text{and}\ \vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big).$
Answer$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$
$\vec{\text{a}}_1=4\hat{\text{i}}-\hat{\text{j}}\ \ \ \vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big)$
$\vec{\text{a}}_2=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}_2=2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\cdot\big(\vec{\text{b}}_2\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\end{vmatrix}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}} &\hat{\text{j}}&\hat{\text{k}}\\1 & 2&-3\\2&4&-5 \end{vmatrix}$
$=\hat{\text{i}}(-10+12)-\hat{\text{j}}(-5+6)+\hat{\text{k}}(4-4)$
$=2\hat{\text{i}}-\hat{\text{j}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(-3\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)\cdot\big(2\hat{\text{i}}-\hat{\text{j}}\big)}{\sqrt{4+1}}\end{vmatrix}$
$=\begin{vmatrix}\frac{-6}{\sqrt{5}}\end{vmatrix}=\frac{6}{\sqrt{5}}$
View full question & answer→Question 1965 Marks
Find the shortest distance between the following two lines:
$\vec{\text{r}}=\text{(1 +}\lambda)\hat{\text{i}}+\text{(2 -}\lambda)\hat{\text{j}}+(\lambda+\text{1)}\hat{\text{k}};$
$\vec{\text{r}}=(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}})+\mu(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}).$
AnswerHere $\vec{\text{a}_{1}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k},}\vec{\text{ b}_{1}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}_{2}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k},}\vec{\text{ b}_{2}}=2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{2k}}$
$\vec{\text{a}_{2}}-\vec{\text{a}_{1}}=\hat{\text{i}}-3{\text{j}}-2{\text{k}}$
$\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}=-3\hat{\text{i}}+3{\text{k}}$
Shortest distance (d) = $\Bigg|\frac{\Big(\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}\Big)\cdot\Big(\vec{\text{a}_{2}}-\vec{\text{a}_{1}}\Big)}{\Big|\vec{\text{b}_{1}}\times\vec{\text{b}_{2}}\Big|}\Bigg|$
=$\Bigg|\frac{\Big(-3\hat{\text{i}}+3\hat{\text{k}}\Big)\cdot\Big(\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\Big)}{\sqrt{9 + 9}}\Bigg|$
$=\frac{9}{3\sqrt{2}}\text{ OR }\frac{3\sqrt{2}}{2}$.
View full question & answer→Question 1975 Marks
Find the vector equation of the line passing through the point (1, -1, 2) and perpendicular to the plane 2x - y + 3z - 5 = 0.
AnswerLet a, b, c be the direction ratios of the given line.
Since the line passes through the point (1, -1, 2) is
$\frac{\text{x}-1}{\text{a}}=\frac{\text{y}+1}{\text{b}}=\frac{\text{z}-2}{\text{c}}\ ...(\text{i})$
$\text{a}=2\lambda,\text{ b}=-\lambda,\text{ c}=3\lambda$
Substituting these values in (i) we get
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}-2}{3},$ which is the cartesian from of the line.
Vector form
The given line passes through a point whose position vector is $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
So, its equation in vector form is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})+\lambda(\hat{2\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$
View full question & answer→Question 1985 Marks
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}=\frac{1}{\text{p}^2}.$
AnswerWe know that equation of plane making intercepts a, b, c (on the axes) is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}-1=0$
Given: Perpendicular distance of the origin (0, 0, 0) from plane = p
$\therefore\ \frac{|\text{a}\text{x}_1+\text{b}\text{y}_1+\text{c}\text{z}_1+\text{d}|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}=\frac{\Big|\frac{0}{\text{a}}+\frac{0}{\text{b}}+\frac{0}{\text{c}}-1\Big|}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}}=\text{p}$
$\Rightarrow\frac{|-1|}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}}=\text{p}$
Squaring both sides, $\frac{1}{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}=\text{p}^2$
$\Rightarrow\ \text{p}^2\Big({\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}\Big)=1$
$\Rightarrow\ \Big({\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}}\Big)=\frac{1}{\text{p}^2}$
View full question & answer→Question 1995 Marks
Show that the points A, B, C with position vectors $2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \hat{\text{i}} - 3\hat{\text{j}} - 5\hat{\text{k}} \text{ and } 3\hat{\text{i}} - 4\hat{\text{j}} - 4\hat{\text{k}}$ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle.
Answer$\vec{\text{AB}} = - \hat{\text{i}} - 2\hat{\text{j}} - 6\hat{\text{k}}, \vec{\text{BC}} = 2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}, \vec{\text{CA}} = -\hat{\text{i}} + 3\hat{\text{j}} + 5\hat{\text{k}}$
Since $\vec{\text{AB}}, \vec{\text{BC}}, \vec{\text{CA}},$ are not parallel vectors, and $\vec{\text{AB}} + \vec{\text{BC}} + \vec{\text{CA}} = \vec{0} \therefore \text{A, B, C}$ form a triangle
$\text{Also} \vec{\text{ BC}}. \vec{\text{CA}} = 0 \text{ }\text{ }\text{ }\text{ }\text{ } \therefore\text{A, B, C}$ form a right triangle
$\text{Area of} \Delta = \frac{1}{2} | \vec{\text{AB}} \times \vec{\text{BC}}| = \frac{1}{2} \sqrt{210}$
View full question & answer→Question 2005 Marks
Find the position vector of the food of perpendicular and the perpendicular distance from the point P with position vector $2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ to the plane $\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})-26=0.$ Also find image or P in the plane.
AnswerLet M be the foot of the perpendicular drawn from the point P(2, 3, 4) in the plane
$\vec{\text{r}}.(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})-26=0$ or 2x + y + 3z - 26 = 0
Then, PM is the normal to the plane. So, the direction rations of PM are proportional to 2, 1, 3.
Since PM passes throught P(2, 3, 4) and has direction rations proportional to 2, 1, 3 so the equation or PM is
$\frac{\text{x}-2}{2}=\frac{\text{y}-3}{1}=\frac{\text{z}-4}{3}=\text{r (say)}$
Let the coordinates or M be (2r + 2, r + 3, 3r + 4). Since M lies in the plane 2x + y + 3z - 26 = 0, So
2(2r + 2) + r + 3 + 3(3r + 4) - 26 = 0
⇒ 4r + 4 + r + 3 + 9r + 12 - 26 = 0
⇒ 14r - 7 = 0
$\Rightarrow\text{r}=\frac{1}{2}$
Therefore, the coordinates of M are
$(2\text{r}+2,\text{r}+3,3\text{r}+4)\\=\Big(2\times\frac{1}{2}+2,\frac{1}{2}+3,3\times\frac{1}{2}+4\Big)\\=\Big(3,\frac{7}{2},\frac{11}{2}\Big)$
Thus, the position vector of the foot of perpendicular are $3\hat{\text{i}}+\frac{7}{2}\hat{\text{j}}+\frac{11}{2}\hat{\text{k}}.$
Now,
Length of the perpendicular from P on to the given plane
$=\Big|\frac{2\times2+1\times3+3\times4-26}{\sqrt{4+1+9}}\Big|$
$=\frac{7}{\sqrt{14}}$
$=\sqrt{\frac{7}{2}}\text{ units}$
Let $Q(x_1, y_1, z_1)$ be the image of point P in the given plane.
Then, the coordinates of M are $\Big(\frac{\text{x}_1+2}{2},\frac{\text{y}_1+3}{2},\frac{\text{z}_1+4}{2}\Big)$
But, the coordinate or M are $\Big(3,\frac{7}{2},\frac{11}{2}\Big)$
$\therefore\Big(\frac{\text{x}_1+2}{2},\frac{\text{y}_1+3}{2},\frac{\text{z}_1+4}{2}\Big)=\Big(3,\frac{7}{2},\frac{11}{2}\Big)$
$\Rightarrow \frac{\text{x}_1+2}{2}=3,\frac{\text{y}_1+3}{2}=\frac{7}{2},\frac{\text{z}_1+4}{2}=\frac{11}{2}$
$\Rightarrow\text{x}_1=4,\text{y}_1=4,\text{z}_1=7$
Thus, the coordinates of the image of the point P in the given plane are (4, 4, 7).
View full question & answer→