Question 513 Marks
Find the distance of the point $2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}$ from the plane $\vec{\text{r}}\cdot(3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}})-9=0$
AnswerWe know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$ is given by
$\text{p}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
Here, $\vec{\text{a}}=-2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}; \vec{\text{n}}=3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}};\text{d}=9$
So, the required distance, p
$=\frac{\big|(2\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}})\cdot(3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}})-9\big|}{\big|3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}}-9\big|}$
$=\frac{6+4-48-9}{\sqrt{9+16+144}}$
$=\frac{|-47|}{13}$
$=\frac{47}{13}\text{ units}$
View full question & answer→Question 523 Marks
Find the equation of the plane passing through (a, b, c) and parallel to the plane $\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=2.$
AnswerEquation of any plane parallel to the plane $\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=2$ is $\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=\lambda\ \ ....(\text{i})$ Plane (i) passes through (a, b, c) $\therefore$ Putting $\vec{\text{r}}=(\text{a},\ \text{b},\ \text{c})=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ in eq. (i), we get $\Big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\Big).\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=\lambda$$\Rightarrow\ \ \text{a}(1)+\text{b}(1)+\text{c}(1)=\lambda$
$\Rightarrow\ \ \lambda=\text{a}+\text{b}+\text{c}$ Putting the value of $\lambda$ in eq. (i), to get the required plane is$\vec{\text{r}}.\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=\text{a}+\text{b}+\text{c}.$
View full question & answer→Question 533 Marks
Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y - z = 1 and 3x - 4y + z = 5.
AnswerThe equation of any plane passing through the origin (0, 0, 0) is,
a(x - 0) + b(y - 0) + c(z - 0) = 0
ax + by + cz = 0 ...(i)
It is given that (i) is perpendicular to the planes x + 2y - z = 1 and 3x - 4y + z = 5. Then,
a + 2b - c = 0 ....(ii)
3a - 4b + c = 0 .....(ii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}&\text{y}&\text{z}\\1&2&-1\\3&-4&1\end{vmatrix}=0$
⇒ -2x - 4y - 10z = 0
⇒ x + 2y + 5z = 0
View full question & answer→Question 543 Marks
Find the equation of a passing through the point (-1, -1, 2) and perpendicular to the planes 3x + 2y - 3z = 1 and 5x - 4y + z = 5.
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 each of the planes 3x + 2y - 3z = 1 and 5x - 4y + z = 5. Then,
3a + 2b - 3c = 0 ....(i)
5a - 4b + c = 0 .....(ii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}+1&\text{y}+1&\text{z}-2\\3&2&-3\\5&-4&1\end{vmatrix}=0$
⇒ -10(x + 1) - 18(y + 1) - 22(z - 2) = 0
⇒ 5(x + 1) + 9(y + 1) + 11(z - 2) = 0
⇒ 5x + 5 + 9y + 9 + 11z - 22 = 0
⇒ 5x + 9y + 11z - 8 = 0
View full question & answer→Question 553 Marks
Find the coordinates of the point where the line
$\frac{\text{x + 1}}{2}=\frac{\text{y + 2}}{3}=\frac{\text{z + 3}}{4}$
meets the plane x + y + 4z = 6.
AnswerAny Point of the line $\frac{\text{x + 1}}{2}=\frac{\text{y + 2}}{3}=\frac{\text{z + 3}}{4}\text{ is (2}\lambda-1,3\lambda-2,4\lambda-3)$If the line meets the plane, then this point must satisfy the equation of plane for some value of $\lambda$
$\therefore(2\lambda-1)+(3\lambda-2)+4(4\lambda-3)=6\text{ }\text{ }\Rightarrow\lambda=1$
$\therefore$Coordinates of required point are (1, 1, 1).
View full question & answer→Question 563 Marks
Find the vector and cartesian equations of the planes:
That passes through the point (1, 0, -2) and the normal to the plane is $\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}.$
AnswerThe position vector of point (1, 0, -2) is $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{k}}$
The normal vector $\vec{\text{N}}$ perpendicular to the plane is $\vec{\text{N}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
The vector equation of the plane is given by, $\Big(\vec{\text{r}}-\vec{\text{a}}\Big).\vec{\text{N}}=0$
$\Rightarrow\Big[\vec{\text{r}}-\Big(\hat{\text{i}}-2\hat{\text{k}}\Big)\Big].\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=0\ \ \ ....(1)$
$\vec{\text{r}}$ is the position vector of any point P(x, y, z) in the plane.
$\therefore\ \ \vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Therefore, equation (1) becomes
$\Big[\Big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\Big)-\Big(\hat{\text{i}}-2\hat{\text{k}}\Big)\Big].\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=0$
$\Rightarrow\Big[(\text{x}-1)\hat{\text{i}}+\text{y}\hat{\text{j}}+(\text{z}+2)\hat{\text{k}}\Big].\Big(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\Big)=0$
⇒ (x - 1) + y - (z + 2) = 0
⇒ x + y - z - 3 = 0
⇒ x + y - z = 3
This is the Cartesian equation of the required plane.
View full question & answer→Question 573 Marks
Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})=9$
AnswerWe know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
$\vec{\text{n}}_2=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\big|2\hat{\text{i}}+3\hat{\text{j}}-6{\hat{\text{k}}\big|}\big|\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big|}$
$=\frac{2-16-12}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{-16}{(7)(3)}$
$=\frac{-16}{21}$
$\theta=\cos^{-1}\Big(\frac{-16}{21}\Big)$
View full question & answer→Question 583 Marks
If $O$ be the origin and the coordinates of P be $(1, 2, -3)$, then find the equation of the plane passing through $P$ and perpendicular to $OP.$
AnswerGiven: Origin $O(0, 0, 0)$ and point $P(1, 2, -3)$
To find: Equation of the plane passing through $P(1, 2, - 3) = (x_1, y_1, z_1)$
$\therefore$ Direction ratios of normal OP to the plane are $1 - 0, 2 - 0, -3 - 0$
$\Rightarrow 1, 2, -3 = (a, b, c)$
$\therefore$ Equation of the required plane is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$
$\Rightarrow 1(x - 1) + 2(y - 2) - 3(z + 3) = 0$
$\Rightarrow x - 1 + 2y - 4 - 3z - 9 = 0$
$\Rightarrow x + 2y - 3z - 14 = 0.$
View full question & answer→Question 593 Marks
Show that the line through the points $(4, 7, 8), (2, 3, 4)$ is parallel to the line through the points $(-1, -2, 1), (1, 2, 5).$
AnswerWe know that direction ratios of the line joining the points $A(4, 7, 8)$ and $B(2, 3, 4)$ are
$x_2 - x_1, y_2 - y_1, z_2 - z_1$
$\Rightarrow 2 - 4, 3 - 7, 4 - 8$
$\Rightarrow -2, -4, -4 = a_1, b_1, c_{1 (say)}$
Again, direction ratios of the line joining the points $C(-1, -2, 1)$ and $D(1, 2, 5)$ are
$x_2 - x_1, y_2 - y_1, z_2 - z_1$
$\Rightarrow 1 - (-1), 2 - (-2), 5 - 1$
$\Rightarrow 2, 4, 4 = a_2, b_2, c_2$ (say)
For lines AB and CD,
$\frac{\text{a}_1}{\text{a}_2}=\frac{-2}{2},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-4}{4},\ \frac{\text{c}_1}{\text{c}_2}=\frac{-4}{4}=-1$
$\therefore\ \ \ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Therefore, line $AB$ is parallel to line $CD.$
View full question & answer→Question 603 Marks
Find the equation of a line parallel to x-axis and passing through the origin.
AnswerWe know that a unit vector along x-axis is $\hat{\text{i}}=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\therefore$ Direction cosines of x-axis are coefficient of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ in the unit vector i.e., 1, 0, 0 = l, m, n
$\therefore$ Equation of the required line passing through the origin (0, 0, 0) and parallel to x-axis is
$\frac{\text{x}-0}{1}=\frac{\text{y}-0}{0}=\frac{\text{z}-0}{0}\ \Rightarrow\ \frac{\text{x}}{1}=\frac{\text{y}}{0}=\frac{\text{z}}{0}$
Vector equation of the required line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\ \ \vec{\text{r}}=\vec{0}+\lambda\hat{\text{i}}\ \ \ \ \ [\vec{\text{a}}=\vec{0}\ \text{and}\ \vec{\text{b}}=\hat{\text{i}}]$
$\Rightarrow\ \ \ \vec{\text{r}}=\lambda\hat{\text{i}}$
View full question & answer→Question 613 Marks
Find the coordinates of the foot of the perpendicular drawn from the origin.
$3y + 4z - 6 = 0$
AnswerLet the coordinates of the foot of perpendicular P from the origin to the plane $b(x_1, y_1, z_1)$.
$3y + 4z - 6 = 0$
$\Rightarrow 0x + 3y + 4z = 6$ ....(1)
The direction ratios of the normal are 0, 3, and 4.
$\therefore\ \ \sqrt{0+3^2+4^2}=5$
Dividing both sides of equation (1) by 5, we obtain
$0\text{x}+\frac{3}{5}\text{y}+\frac{4}{5}\text{z}=\frac{6}{5}$
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of
normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Therefore, the coordinates of the perpendicular are
$\Big(0,\ \frac{3}{5}.\frac{6}{5},\ \frac{4}{5}.\frac{6}{5}\Big)\ \text{i.e.},\ \Big(0,\ \frac{18}{25},\ \frac{24}{25}\Big).$
View full question & answer→Question 623 Marks
Find the equation of the plane which passes through the point (3, 4, -1) and is parallel to the plane 2x - 3y + 5z + 7 = 0. Also, find the distance between the two planes.
AnswerEquation of plane is parallel to 2x - 3y + 5z + 7 = 0 is of the form 2x - 3y + 5z = d
Above plane is passing through (3, 4, -1)
So, substitute above point in the equation, we get
6 - 12 - 5 = d
d = -11
So, palne equation is 2x - 3y + 5z = -11
Distance between planes is given by
$\Big|\frac{-7+11}{\sqrt{4+9+25}}\Big|=\frac{4}{\sqrt{38}}$
View full question & answer→Question 633 Marks
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes $\vec{\text{r}}.\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)=5\ \text{and}\ \vec{\text{r}}.\Big(3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)=6.$
AnswerThe required line passes through the point $\text{A}(1, 2, 3) =\vec{\text{a}}$
$\therefore\ \vec{\text{a}}=\text{Position vector of point A}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Let $\vec{\text{b}}$ be any vector along the required line.
$\therefore$ Vector equation of required line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\ \vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda\vec{\text{b}}\ \ \ ...(\text{i})$
Since required line is parallel to the plane $\vec{\text{r}}.\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)=5$
$\therefore\ \vec{\text{b}}.\vec{\text{n}_1}=0\ \text{and}\ \vec{\text{b}}.\vec{\text{n}_2}=0$
Comparing with $\vec{\text{r}}.\vec{\text{n}_1}=\text{d}_1$ we have, $\vec{\text{n}_1}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
And Comparing with $\vec{\text{r}}.\vec{\text{n}_2}=\text{d}_2$ we have, $\vec{\text{n}_2}=3\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Since $\vec{\text{b}}$ isperpendicular to both $\vec{\text{n}_1}\ \text{and}\ \vec{\text{n}_2}$
$\therefore\ \vec{\text{b}}=\vec{\text{n}_1}\times\vec{\text{n}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-1&2\\3&1&1\end{vmatrix}$
Expanding along first row,
$\vec{\text{b}}=\hat{\text{i}}(-1-2)-\hat{\text{j}}(1-6)+\hat{\text{k}}(1+3)=-3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
Putting this value of $\vec{\text{b}}$ in eq. (i), vector equation of required line,
$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda\Big(-3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}\Big).$
View full question & answer→Question 643 Marks
Write the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=14$ in normal form.
AnswerThe given equation of the plane is
$\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}})=14$ or $\vec{\text{r}}.\vec{\text{n}}=14$, where $\vec{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{4+9+36}=7$
For reducing the given equation to normal form, we need to divide it by $|\vec{\text{n}}|$.
Then, we get $\vec{\text{r}}.\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{14}{|\vec{\text{n}}|}$
$=\vec{\text{r}}.\Big(\frac{2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}}{7}\Big)=\frac{14}{7}$
$\Rightarrow\vec{\text{r}}.\Big(\frac{2}{7}\hat{\text{i}}+\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=2,$ which is the required normal form.
View full question & answer→Question 653 Marks
Find the angle between the given planes.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})=1$ and $\vec{\text{r}}\cdot(-\hat{\text{i}}+\hat{\text{j}})=4$
AnswerWe know that the angle between the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ is given by
$\cos\theta=\frac{\vec{\text{n}}_1\cdot\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{n}}_2=-\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
So, $\cos\theta=\frac{(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}})\cdot(-\hat{\text{i}}+\vec{\text{j}}+0\hat{\text{k}})}{\big|2\hat{\text{i}}-3\hat{\text{j}}+4{\hat{\text{k}}\big|}\big|\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big|}$
$=\frac{-2-3}{\sqrt{4+9+16}\sqrt{1+1+0}}$
$=\frac{-5}{\sqrt{29}\sqrt{2}}$
$=\frac{-5}{\sqrt{58}}$
$\theta=\cos^{-1}\Big(\frac{-5}{\sqrt{58}}\Big)$
View full question & answer→Question 663 Marks
A plabne makes intercepts -6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.
AnswerWe know that the equation of the plane which makes intercepts a, b and c on the coordinate axes is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
So, the equation of the plane which makes intercepts -6, 3, 4 on the x-axis, y-axis and z-axis, respecticely is,
$\frac{\text{x}}{-6}+\frac{\text{y}}{3}+\frac{\text{z}}{4}=1$
$\Rightarrow-2\text{x}+4\text{y}+3\text{z}=12$
$\Rightarrow-2\text{x}+4\text{y}+3\text{z}+12=0$
$\therefore$ Length of the perpendicular from (0, 0, 0) to the plane 2x - 4y - 3z + 12 = 0
$=\Bigg|\frac{2\times0-4\times0-3\times0+12}{\sqrt{12^2+(-4)^2+(-3)^2}}\Bigg|$
$=\bigg|\frac{12}{\sqrt{4+16+9}}\bigg|$
$=\frac{12}{\sqrt{29}}\text{ units}$
Thus, the length of the perpendicular from the origin to the plane is $\frac{12}{\sqrt{29}}\text{ units}.$
View full question & answer→Question 673 Marks
Find the equation of the plane passing through (a, b, c) and parallel to the plane $\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=2$
AnswerThe equation of the family of plane parallel to $\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=2$ is,
$\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=\text{d}\ ...(\text{i})$
If it passes through (a, b, c) then
$(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=\text{d}$
$=\text{a}+\text{b}+\text{c}$
Substituting a + b + c = d in (i) we get
$\vec{\text{r}}\cdot(\text{i}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}$
$\text{x}+\text{y}+\text{z}=\text{a}+\text{b}+\text{c}$ as the equation of the required plane.
View full question & answer→Question 683 Marks
A line passes throuth the point with position vector $2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and is in the direction of $3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}.$ Find equations of the line in vector and cartesian form.
AnswerWe know that the vector equation of a line passing through a point with position vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}.$
Here,
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
So, the vector equation of the required line is
$\vec{\text{r}}=\big(2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+\lambda\big(3\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.
View full question & answer→Question 693 Marks
Show that the points $(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$ and $3(\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})$ are equidistant from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$ and lies on opposite side of it.
AnswerTo show that these given points $(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$ and $3(\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})$ are equidistant from the plane
$\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0,$ we first find out the mid-point of the points which is $(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}).$
On substituting $\vec{\text{r}}$ by the mid-point in plane, we get
$\text{L.H.S.}=(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})\cdot(5\hat{\text{i}}+\hat{\text{j}}-7\hat{\text{k}})+9$
$=10+2-21+9=0$
$=\text{R.H.S.}$
Hence, the two points lie on opposite sides of the plane are equidistant from the plane.
View full question & answer→Question 703 Marks
Find a unit normal vector to the plane x + 2y + 3z - 6 = 0
AnswerThe given equation of the plane is
x + 2y + 3z - 6 = 0
x + 2y + 3z = 6
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=6$
Or $\vec{\text{r}}\cdot\vec{\text{n}}=6$
When, $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\ ...(\text{i})$
Now, $|\vec{\text{n}}|=\sqrt{1^2+2^2+3^2}$
$=\sqrt{1+4+9}$
$=\sqrt{14}$
Unit vector to the plane, $\hat{\text{n}}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}$
$=\frac{\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}}{\sqrt{14}}$
$=\frac{1}{\sqrt{14}}\hat{\text{i}}+\frac{2}{\sqrt{14}}\hat{\text{j}}+\frac{3}{\sqrt{14}}\hat{\text{k}}$
View full question & answer→Question 713 Marks
Find the angle between the following pairs of lines:
- $\vec{\text{r}}=3\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}+\lambda\Big(\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}\Big)\ \text{and}$
$\vec{\text{r}}=2\hat{\text{i}}-\hat{\text{j}}-56\hat{\text{k}}+\mu\Big(3\hat{\text{i}}-5\hat{\text{j}}-4\hat{\text{k}}\Big)$ Answer
- Comparing the first and second equation with $\Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)\ \text{and}\ \Big(\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{b}}\Big)$ resp.
$ \vec{\text{b}_1}=\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=3\hat{\text{i}}-5\hat{\text{j}}-4\hat{\text{k}}$
Let $\theta$ be the angle between these two lines, then
$\cos\theta=\frac{\vec{\text{b}_1}.\vec{\text{b}_2}}{\Big|\vec{\text{b}_1}\Big|.\Big|\vec{\text{b}_2}\Big|} =\frac{1(3)+(-1)(-5)+(-2)(-4)}{\sqrt{1+1+4}\sqrt{9+25+16}}=\frac{3+5+8}{\sqrt{6}\sqrt{50}}=\frac{16}{\sqrt{300}}$
$\cos\theta=\frac{16}{10\sqrt{3}}=\frac{8}{5\sqrt{3}}\ \ \ \ \ \ \ \Rightarrow\ \theta=\cos^{-1}\frac{8}{5\sqrt{3}}.$ View full question & answer→Question 723 Marks
If a line has the direction ratios -18, 12, -4, then what are its direction cosines?
AnswerWe know that if a, b, c are direction ratios of a line, then direction cosines of the line are:
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}\ \ \ ....(\text{i})$
Here direction ratios of the line are -18, 12, -4
Putting the values in eq. (i),
$\frac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}},\ \frac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}},\ \frac{\text{-4}}{\sqrt{(-18)^2+(12)^2+(-4)^2}}$
$\Rightarrow\ \frac{-18}{\sqrt{324+144+16}},\ \frac{12}{\sqrt{324+144+16}},\ \frac{\text{-4}}{\sqrt{324+144+16}}$
$\Rightarrow\ \frac{-18}{\sqrt{484}},\ \frac{12}{\sqrt{484}},\ \frac{\text{-4}}{\sqrt{484}}\ \Rightarrow\ \frac{-18}{22},\ \frac{12}{22},\ \frac{-4}{22}$
$\Rightarrow\ \frac{-9}{11},\ \frac{6}{11},\ \frac{-2}{11}$
Hence, direction cosines of required lines are $\frac{-9}{11},\ \frac{6}{11},\ \frac{-2}{11}.$
View full question & answer→Question 733 Marks
Show that the line through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).
AnswerSuppose vector $\vec{\text{a}}$ is passing through the points (1, -1, 2) and (3, 4, -2) and $\vec{\text{b}}$ passing through the points (0, 3, 2) and (3, 5, 6).
Then,
$\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
Now,
$\vec{\text{a}}.\vec{\text{b}}=\big(2\hat{\text{i}}+5\hat{\text{j}}-4\hat{\text{k}}\big).\big(3\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)=0$
Hence, the given lines are perpendicular to each other.
View full question & answer→Question 743 Marks
Find the vector equation of the plane with intercepts 3, -4 and 2 on x, y and z-axis respectively.
AnswerThe equation of the plane in the intercept form is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1,$ where a, b and c are the intercepts on the x, y and z-axis, respectively.
It is given that intercepts made by the plane on the X, Y and Z-axis are 3, -4 and 2, respectively.
$\therefore$ a = 3, b = -4, c = 2
Thus, the equation of the plane is
$\frac{\text{x}}{3}+\frac{\text{y}}{(-4)}+\frac{\text{z}}{2}=1$
$\Rightarrow4\text{x}-3\text{y}+6\text{z}=12$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(4\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}})=12$
This is the vector form of the equation of the given plane.
View full question & answer→Question 753 Marks
Find the coordinates of the foot of the perpendicular drawn from the origin.
$x + y + z = 1$
AnswerLet the coordinates of the foot of perpendicular P from the origin to the plane be $(x_1, y_1, z_1)$.
$x + y + z = 1$
The direction ratios of the normal are 1, 1, and 1.
$\therefore\ \ \sqrt{1^2+1^2+1^2}=\sqrt{3}$
Dividing both sides of equation (1) by $\sqrt{3},$ we obtain
$\frac{1}{\sqrt{3}}\text{x}+\frac{1}{\sqrt{3}}\text{y}+\frac{1}{\sqrt{3}}\text{z}=\frac{1}{\sqrt{3}}$
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of
normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Therefore, the coordinates of the foot of the perpendicular are
$\Big(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}.\frac{1}{\sqrt{3}}\Big)\ \text{i.e.},\ \Big(\frac{1}{{3}},\ \frac{1}{{3}},\ \frac{1}{{3}}\Big).$
View full question & answer→Question 763 Marks
Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.
AnswerSince, the equation of a plane is bisecting perpendicular the line joining the points A (2, 3, 4) and B (4, 5, 8) at right angles.
So, mid-point of AB is $\Big(\frac{2+4}{2},\frac{3+5}{2},\frac{4+8}{2}\Big)$ i.e., (3, 4, 6)
Also, normal to the plane, $\vec{\text{N}}=(4-2)\hat{\text{i}}+(5-3)\hat{\text{j}}+(8-4)\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
So, the required equation of the plane is $(\vec{\text{r}}-\vec{\text{a}})\cdot\vec{\text{N}}=0 $
$\Rightarrow\Big[(\text{x}-3)\vec{\text{i}}+(\text{y}-4)\vec{\text{j}}(\text{z}-6)\vec{\text{k}}\Big]\cdot(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}})=0$ $\Big[\because\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}\Big]$
$\Rightarrow2\text{x}-6+2\text{y}-8+4\text{z}-24=0$
$\Rightarrow2\text{x}+2\text{y}+4\text{z}=38$
$\Rightarrow\text{x}+\text{y}+2\text{z}=19$
View full question & answer→Question 773 Marks
Reduce the equation of the following planes to intercept from and find the intercepts on the coordinate axes:
2x - y + z = 5
AnswerEquation of the given plane is,2x - y + z = 5
Dividng both sides by 5, we get
$\frac{2\text{x}}{5}+\frac{-\text{y}}{5}+\frac{\text{z}}{5}=\frac{5}{5}$
$\Rightarrow\frac{\text{x}}{\big(\frac{5}{2}\big)}+\frac{\text{y}}{-5}+\frac{\text{z}}{5}=1\ ...(\text{i})$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$
Comparing (i) and (ii) we get
$\text{a}=\frac{5}{2};\text{ b}=-5;\text{ c}=5$
View full question & answer→Question 783 Marks
O is the origin and A is (a, b, c).Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.
AnswerHere, Direction Cosines of line OA are $\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$ and $\frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
Also $\vec{\text{n}}=\overrightarrow{\text{OA}}$
$=\vec{\text{a}}=\text{a}\vec{\text{i}}+\text{b}\vec{\text{j}}+\text{c}\vec{\text{k}}$
The equation of plane passes through (a, b, c) and perpendicular to OA is given by
$\big[\vec{\text{r}}-\vec{\text{a}}\big]\cdot\vec{\text{n}}=0$
$\Rightarrow\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\Big[(\text{x}\vec{\text{i}}+\text{y}\vec{\text{j}}+\text{z}\vec{\text{k}})\cdot(\text{a}\vec{\text{i}}+\text{b}\vec{\text{j}}+\text{c}\vec{\text{k}})\Big]$ $=(\text{a}\vec{\text{i}}+\text{b}\vec{\text{j}}+\text{c}\vec{\text{k}})\cdot({\text{a}\vec{\text{i}}+\text{b}\vec{\text{j}}+\text{c}\vec{\text{k}}})$
$\Rightarrow\text{ax}+\text{by}+\text{cz}=\text{a}^2+\text{b}^2+\text{c}^2$
View full question & answer→Question 793 Marks
Find the angle between the following pair of lines:
- $\frac{\text{x}-2}{2}=\frac{\text{y}-1}{5}=\frac{\text{z}+3}{-3}\ \text{and}\ \frac{\text{x}+2}{-1}=\frac{\text{y}-4}{8}=\frac{\text{z}-5}{4}$
Answer
- Given: Equation of first line is $\frac{\text{x}-2}{2}=\frac{\text{y}-1}{5}=\frac{\text{z}+3}{-3}$
The direction ratios of this line i.e., a vector along the line is
$\vec{\text{b}_1}=(2,\ 5,-3)=2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}}$
Now equation of second line is $\frac{\text{x}+2}{-1}=\frac{\text{y}-4}{8}=\frac{\text{z}-5}{4}$
The direction ratios of this line i.e., a vector along the line is
$\vec{\text{b}_2}=(-1,\ 8,\ 4)=-\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}$
Let $\theta$ be the angle between these two lines, then
$\cos\theta=\frac{\vec{\text{b}_1}.\vec{\text{b}_2}}{\Big|\vec{\text{b}_1}\Big|.\Big|\vec{\text{b}_2}\Big|} =\frac{2(-1)+(5)(8)+(-3)(4)}{\sqrt{4+25+9}\sqrt{1+64+16}}=\frac{-2+40-12}{\sqrt{38}\sqrt{81}}=\frac{26}{9\sqrt{38}}$
$\Rightarrow\ \ \ \theta=\cos^{-1}\frac{26}{9\sqrt{38}}.$ View full question & answer→Question 803 Marks
Find the coordinates of the foot of the perpendicular drawn from the origin.
$5y + 8 = 0$.
AnswerLet the coordinates of the foot of perpendicular P from the origin to the plane be $(x_1, y_1, z_1)$.
$5y + 8 = 0$
$⇒ 0x - 5y + 0z = 8 ....(1)$
The direction ratios of the normal are 0, -5, and 0.
$\therefore\ \ \sqrt{0+(-5)^2+0}=5$
Dividing both sides of equation (1) by 5, we obtain
$-\text{y}=\frac{8}{5}$
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of
normal to the plane and d is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by (ld, md, nd).
Therefore, the coordinates of the foot of the perpendicular are
$\Big(0,\ -1\Big(\frac{8}{5}\Big),\ 0\Big)\ \text{i.e.}\ \Big(0,\ -\frac{8}{5},\ 0\Big).$
View full question & answer→Question 813 Marks
Find the angle between the planes whose vector equations are:
$\vec{\text{r}}.\Big(2\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)=5\ \text{and}\ \vec{\text{r}}.\Big(3\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\Big)=3.$
AnswerThe equations of the given planes are
$\vec{\text{r}}.\Big(2\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)=5\ \text{and}\ \vec{\text{r}}.\Big(3\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\Big)=3$
It is known that if $\vec{\text{n}}_1\ \text{and}\ \vec{\text{n}}_2$ are normal to the planes, $\vec{\text{r}}.\vec{\text{n}}_1=\vec{\text{d}}_1\ \text{and}\ \vec{\text{r}}.\vec{\text{n}}_2=\vec{\text{d}}_2,$ then the angle between them, Q is given by,
$\cos\text{Q}=\Bigg|\frac{\vec{\text{n}}_1.\vec{\text{n}}_2}{|\vec{\text{n}}_1||\vec{\text{n}}_2|}\Bigg|\ \ ...(1)$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\ \text{and }\vec{\text{n}}_2=3\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}$
$\therefore\ \ \vec{\text{n}}_1.\vec{\text{n}}_2=\Big(2\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)\Big( 3\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\Big)$
= 2.3 + 2.(-3) + (-3).5 = -15
$|\vec{\text{n}}_1|=\sqrt{(2)^2+(2)^2+(-3)^2}=\sqrt{17}$
$|\vec{\text{n}}_2|=\sqrt{(3)^2+(-3)^2+(5)^2}=\sqrt{43}$
$\cos\text{Q}=\Big|\frac{-15}{\sqrt{17}.\sqrt{43}}\Big|$
$\Rightarrow\ \cos\text{Q}=\frac{15}{\sqrt{731}}$
$\Rightarrow\ \text{Q}=\cos^{-1}\Big(\frac{15}{\sqrt{731}}\Big).$
View full question & answer→Question 823 Marks
Find the vector equation of the line passing through the point A(1, 2, -1) and parallel to the line 5x - 25 = 14 - 7y = 35z.
AnswerThe equation of the line 5x - 25 = 14 - 7y = 35z can be re-written as
$\frac{\text{x}-5}{\frac{1}{5}}=\frac{\text{y}-2}{\frac{-1}{7}}=\frac{\text{z}}{\frac{1}{35}}$
$\Rightarrow\frac{\text{x}-5}{7}=\frac{\text{y}-2}{-5}=\frac{\text{z}}{1}$
Since the required line is parallel to the given line, so the direction ration of the required line are proportional to 7, -5, 1.
The vector equation of the required line passing through the point (1, 2, -1) and having direction ratios proportional to 7, -5, 1 is
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(7\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big).$
View full question & answer→Question 833 Marks
Find the vector equation of the line passing through the point (2, -1, -1) which is parallel to the line 6x - 2 = 3y +1 =2z - 2.
AnswerThe equation of the line 6x - 2 = 3y + 1 = 2z - 2 can be re-written as
$\frac{\text{x}-\frac{1}{3}}{\frac{1}{6}}=\frac{\text{y}+\frac{1}{3}}{\frac{1}{3}}=\frac{\text{z}-1}{\frac{1}{2}}$
$=\frac{\text{x}-\frac{1}{3}}{1}=\frac{\text{y}+\frac{1}{3}}{2}=\frac{\text{z}-1}{3}$
Since the reqired line is parallel to the given line, the direction ratios of the recuired line are proportional to 1, 2, 3.
The vector equation of the required line passing through the point (2, -1, -1) and having direction ratios proportional to 1, 2, 3 is $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big).$
View full question & answer→Question 843 Marks
Find the direction cosines of a line which makes equal angles with the coordinate axes.
AnswerLet a line make equal angles $\alpha,\ \alpha,\ \alpha$ with the co-ordinate axes.
$\therefore$ Direction cosines of the line are $\cos\alpha,\ \cos\alpha,\ \cos\alpha\ \ .....(\text{i})$
$\therefore\ \cos^2\alpha+\cos^2\alpha+\cos^2\alpha=1\ \ \ [\because\ \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1]$
$\Rightarrow\ 3\cos^2\alpha\ \Rightarrow\ \cos^2\alpha=\frac{1}{3}\ \Rightarrow\ \cos\alpha=\pm\frac{1}{\sqrt{3}}$
Putting $\cos\alpha=\pm\frac{1}{\sqrt{3}}$ in eq.(i), direction cosines of the required line making equal angles with the co-ordinator axes are $\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}},\pm\frac{1}{\sqrt{3}}$
Direction cosines of a line making equal angles with the co-ordinate axes in the positive i.e., first octant are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
View full question & answer→Question 853 Marks
Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{\text{x}+3}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+8}{6}.$
AnswerGiven: A point on the line is $(-2, 4, -5) = (x_1, y_1, z_1)$
Equation of the given line in Cartesian form is $\frac{\text{x}+3}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+8}{6}$
$\therefore$ Direction ratios of the given line are its denominators $3, 5, 6 = a, b, c$
$\therefore$ Equation of the required line is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{b}=\frac{\text{z}-\text{z}_1}{c}$
$\Rightarrow\ \ \ \frac{\text{x}-(-2)}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}-(-5)}{6}=\frac{\text{x}+2}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+5}{6}$
View full question & answer→Question 863 Marks
If the lines $\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{2\text{k}}=\frac{\text{z}-3}{2}\ \text{and}\ \frac{\text{x}-1}{3\text{k}}=\frac{\text{y}-1}{1}=\frac{\text{z}-6}{-5}$ are perpendicular, find the value of k.
AnswerThe given lines are
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{2\text{k}}=\frac{\text{z}-3}{3}$
and $\frac{\text{x}-1}{3\text{k}}=\frac{\text{y}-1}{1}=\frac{\text{z}-6}{-5}$
Direction ratios of two lines are -3, 2k, 2 and 3k, 1, -5
Since the lines are perpendicular
$\therefore$ (-3)(3k) + (2k)(1) + (2)(-5) = 0
$\therefore$ -9k + 2k - 10 = 0
⇒ -7k = 10
$\therefore\ \text{k}=-\frac{10}{7}$
View full question & answer→Question 873 Marks
Find the coordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ to the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}.$ Hence, of otherwise, deduce the length of the perpendicular.
AnswerLet M be the foot of the perpendicular of the point P(5, 4, 2) on the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}$
Therefore, its equation is
$\frac{\text{x}+1}{2}=\frac{\text{y}-3}{2}=\frac{\text{z}-1}{-1}=\text{r}$
Then, M is in the form $(2r - 1, 3r + 3, -r + 1)$
Direction ratios of MP are $2r - 1 - 5, 3r + 3 - 4, -r + 1 - 2 or 2r - 6, 3r - 1, -r - 1.$
Since MP is perpendicular to the given line $(2, 3, -1), 2 (2r - 6) + 3(3r - 1) -1(-r - 1) = 0$ (Because $a_1, a_2, + b_1, b_2, + c_1, c_2, = 0)$
$\Rightarrow 4r - 12 + 9r - 3 + 1 + 1 = 0$
$\Rightarrow 14r - 14 = 0$
$\Rightarrow r = 1$
So, $M = (2r - 1, 3r + 3, -r + 1) = (2 (1) - 1, 3(1) + 3, -1 + 1) = (1, 6, 0)$
Length of the perperndicular,
$\text{MP}=\sqrt{(1-5)^2+(6-4)^2+(0-2)^2}\\=\sqrt{16+4+4}=\sqrt{24}=2\sqrt{6}\text{ units}$
View full question & answer→Question 883 Marks
Determine the value of $\lambda$ for which the following planes are perpendicular to other.
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})=7$ and $\vec{\text{r}}\cdot(\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=26$
AnswerWe know that the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ are perpendicular to each other only if $\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
Here, $\vec{\text{n}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}, \vec{\text{n}}_2=\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}}$
The given planes are perpendicular.
$\Rightarrow\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
$\Rightarrow(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})\cdot(\lambda\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=0$
$\Rightarrow\lambda+4-21=0$
$\Rightarrow\lambda-17=0$
$\Rightarrow\lambda=17$
View full question & answer→Question 893 Marks
Reduce the equation 2x - 3y - 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.
AnswerGiven equation of plane is,
2x - 3y - 6z = 14
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}})=14$
Dividing the equation by $\sqrt{(2)^2+(-3)^2+(-6)^2}$
$\vec{\text{r}}\cdot\frac{(2\hat{\text{i}}-3\hat{\text{j}}-6\hat{\text{k}})}{\sqrt{4+9+36}}=\frac{14}{\sqrt{4+9+36}}$
$\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=\frac{14}{7}$
$\vec{\text{r}}\Big(\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}\Big)=2\ ...(\text{i})$
We know that the vector equation of a plane with distance d from origin and normal to unit vector $\hat{\text{n}}$ is given by
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{d}}\ ...(\text{ii})$
Comparing (i) and (ii),
d = 2 and
$\hat{\text{n}}=\frac{2}{7}\hat{\text{i}}-\frac{3}{7}\hat{\text{j}}-\frac{6}{7}\hat{\text{k}}$
So, distance of plane from origin = 2 unit
Direction cosine of normal to plane $=\frac{2}{7},-\frac{3}{7},\frac{6}{7}$
View full question & answer→Question 903 Marks
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the:
zx-plane
AnswerDirection ratios of the given line are
(5 - 3, 1 - 4, 6 - 1) = (2, -3, 5)
Hence, equation of the line is
$\frac{\text{x}-5}{2}=\frac{\text{y}-1}{-3}=\frac{\text{z}-6}{5}=\text{r}$
$\Rightarrow\text{x}=2\text{r}+5,\text{ y}=-3\text{r}+1,\text{ z}=5\text{r}+6$
For any point on the zx-plane y = 0
$\Rightarrow-3\text{r}+1=0$
$\Rightarrow\text{r}=\frac{1}{3}$
$\Rightarrow\text{x}=2\Big(\frac{1}{3}\Big)+5=\frac{17}{3}$
$\Rightarrow\text{z}=5\Big(\frac{1}{3}\Big)+6=\frac{23}{3}$
Hence, the corrdinates of the point are $\Big(\frac{17}{3},0,\frac{23}{3}\Big)$
View full question & answer→Question 913 Marks
Write the ratio in which the plane $4x + 5y − 3z = 8$ divides the line segment joining the points $(−2, 1, 5)$ and $(3, 3, 2)$.
AnswerWe know that the ratio in which the plane $ax + by + cz + d = 0$ divides the line sebment joining
$(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\frac{-(\text{ax}_1+\text{by}_1+\text{cz}_1+\text{d})}{\text{ax}_2+\text{by}_2+\text{cz}_2+\text{d}}$
Here, $a = 4,b = 5,c = -3,d = -8,x_1 = -2,y_1 = 1,z_1 = 5,x_2 = 3,y_2 = 3,z_2 = 2$
So, the required ratio
$=\frac{-(4(-2)+5(1)-3(5)-8)}{4(3)+5(3)-3(2)-8}$
$=\frac{-(-8+5-15-8)}{12+15-6-8}$
$=\frac{26}{13}$
$=\frac{2}{1}$ or $2 :1.$
View full question & answer→Question 923 Marks
Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
AnswerA point on the required line is $A(1, 2, 3) = x_1, y_1, z_1$
⇒ Positive vector of a point on the required line is
$\vec{\text{a}}=\overrightarrow{\text{OA}}=(1,2,3)=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
The required line is parallel to the vector $\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}-2\hat{\text{k}}}$
$\therefore$ direction ratios of the required line are coefficient of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}\ \text{in}\ \vec{\text{b}}$ are 3, 2, -2 = a, b, c
$\therefore$ Vector equation of the required line is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow\ \vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\Big)+\lambda\Big(3\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\Big)$
Where $\lambda$ is a real number.
Cartesian equation of this equation is $\frac{\text{x}-1}{3}=\frac{\text{y}-2}{2}=\frac{\text{z}-3}{-2}.$
View full question & answer→Question 933 Marks
Find the direction cosines of the unit vector perpendicular to the plane $\vec{\text{r}}.\big(6\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)+1=0$ passing through the origin.
AnswerGiven equation of the plane $\vec{\text{r}}.\big(6\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)+1=0$
Thus, the direction ratios normal to the plane are 6, -3 and -2
Hence, the direction cosines to the normal to the plane are
$=\frac{6}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-3}{\sqrt{6^2+(-3)^2+(-2)^2}},\frac{-2}{\sqrt{6^2+(-3)^2+(-2)^2}}$
$=\frac{6}{7},\frac{-3}{7},\frac{-2}{7}$
$=\frac{-6}{7},\frac{3}{7},\frac{2}{7}$
The direction cosines of the unit vector perpendicular to the plane are same as the direction cosines of the normal to the plane.
Thus, the direction cosined of the unit vector perpendicular to the plane are: $\frac{-6}{7},\frac{3}{7},\frac{2}{7}$
View full question & answer→Question 943 Marks
Show that the three lines with direction cosines $\frac{12}{13},\ \frac{-3}{13},\ \frac{-4}{13},\ \frac{4}{13},\ \frac{12}{13},\ \frac{3}{13},\ \frac{3}{13},\ \frac{-4}{13},\ \frac{12}{13}$ are mutually perpendicular.
AnswerGiven: Direction cosines of three lines are
$\frac{12}{13},\ \frac{-3}{13},\ \frac{-4}{13}=\text{l}_1,\ \text{m}_1,\ \text{n}_1,$
$\frac{4}{13},\ \frac{12}{13},\ \frac{3}{13}=\text{l}_2,\ \text{m}_2,\ \text{n}_2,$
$\frac{3}{13},\ \frac{-4}{13},\ \frac{12}{13}=\text{l}_3,\ \text{m}_3,\ \text{n}_3,$
For first two lines,
$\text{l}_1\text{l}_2+\text{m}_1\text{m}_2+\text{n}_1\text{n}_2=\Big(\frac{12}{13}\Big)\Big(\frac{4}{13}\Big)+\Big(\frac{-3}{13}\Big)\Big(\frac{12}{13}\Big)+\Big(\frac{-4}{13}\Big)\Big(\frac{3}{13}\Big)$
$=\frac{48}{169}-\frac{36}{169}-\frac{12}{169}=\frac{48-36-12}{169}=\frac{0}{169}=0$
Therefore, the first two lines are perpendicular to each other.
For second and third lines,
$\text{l}_2\text{l}_3+\text{m}_2\text{m}_3+\text{n}_2\text{n}_3=\Big(\frac{4}{13}\Big)\Big(\frac{3}{13}\Big)+\Big(\frac{12}{13}\Big)\Big(\frac{-4}{13}\Big)+\Big(\frac{3}{13}\Big)\Big(\frac{12}{13}\Big)$
$=\frac{12}{169}-\frac{48}{169}+\frac{36}{169}=\frac{12-48+36}{169}=\frac{0}{169}=0$
Therefore, second and third lines are perpendicular to each other.
For First and third lines,
$\text{l}_1\text{l}_3+\text{m}_1\text{m}_3+\text{n}_1\text{n}_3=\Big(\frac{12}{13}\Big)\Big(\frac{3}{13}\Big)+\Big(\frac{-3}{13}\Big)\Big(\frac{-4}{13}\Big)+\Big(\frac{-4}{13}\Big)\Big(\frac{12}{13}\Big)$
$=\frac{36}{169}+\frac{12}{169}-\frac{48}{169}=\frac{36+12-48}{169}=\frac{0}{169}=0$
Therefore, first and third lines are also perpendicular to each other.
Hence, given three lines are mutually perpendicular to each other.
View full question & answer→Question 953 Marks
Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.
AnswerThe given equation of the plane is
2x − 3y + 4z = 12
Dividing both sides by 12, we get
$\Rightarrow\frac{2\text{x}}{\text{12}}+\frac{-3\text{y}}{\text{12}}+\frac{4\text{z}}{\text{12}}=\frac{12}{12}$
$\Rightarrow\frac{\text{x}}{\text{6}}+\frac{\text{y}}{-\text{4}}+\frac{\text{z}}{\text{3}}=1\ ....(1)$
We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(2)$
Comparing (1) and (2), we get
a = 6, b = -4 and c = 3.
View full question & answer→Question 963 Marks
Find the angle between the lines whose direction ratios are a, b, c and b - c, c - a, a - b.
AnswerDirection ratios of one line are a, b, c
⇒ A vector along this line is $\vec{\text{b}_1}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
Direction ratios of second line are b - c, c - a, a - b
⇒ A vector along second line is $\vec{\text{b}_2}=(\text{b - c})\hat{\text{i}}+(\text{c - a})\hat{\text{j}}+(\text{a - b})\hat{\text{k}}$
Let $\theta$ be the angle between the two lines, then
$\cos\theta=\frac{\big|\vec{\text{b}_1}.\vec{\text{b}_2}\big|}{\big|\vec{\text{b}_1}\big|.\big|\vec{\text{b}_2}\big|}=\frac{\text{a}(\text{b - c})+\text{b}(\text{c - a})+\text{c}(\text{a - b})}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{(\text{b - c})^2+(\text{c - a})^2+(\text{a - b})^2}}$
$=\frac{\text{ab - ac}+\text{bc - ab}+\text{ac - bc}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{(\text{b - c})^2+(\text{c - a})^2+(\text{a - b})^2}}=0=\cos90^{\circ}$
$\Rightarrow\ \ \theta=90^{\circ}$
View full question & answer→Question 973 Marks
Show that the following planes are at right angles.
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=3$
AnswerWe know that the planes $\vec{\text{r}}\cdot\vec{\text{n}}_1=\text{d}_1,\vec{\text{ r}}\cdot\vec{\text{n}}_2=\text{d}_2$ are perpendicular to each other only if $\vec{\text{n}}_1\cdot\vec{\text{n}}_2=0$
Here, $\vec{\text{n}}_1=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}},\vec{\text{n}}_2=-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
Now,
$\vec{\text{n}}_1\cdot\vec{\text{n}}_2=\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)\cdot\big(-\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$=-2+1+1=0$
So, the given planes are perpendicular.
View full question & answer→Question 983 Marks
Find the equation of the plane passing throught the point (2, 4, 6) and making equal intercepts on the coordinate axes.
AnswerIntercepts on the coordinate axes are equal.
We know that, if a, b, c are Intercepts on coordinate axes by a plane, then equationb of the plane is given by,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
Here, it is given that a = b = c = p (say)
$\frac{\text{x}}{\text{p}}+\frac{\text{y}}{\text{p}}+\frac{\text{z}}{\text{p}}=1$
$\frac{\text{x}+\text{y}+\text{z}}{\text{p}}=1$
$\text{x}+\text{y}+\text{z}=\text{p}\ ...(\text{i})$
It is given that plane is passing through the point (2, 4, 6), so using equation (i)
x + y + z = p
2 + 4 + 6 = p
12 = p
Put, value of p in equation (i)
x + y + z = 12
So, the required equation of the plane is given by,
x + y + z = 12
View full question & answer→Question 993 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.
$2x + y + 3z - 2$ and $x - 2y + 5 = 0$
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 planes are $2x + y + 3z - 2 = 0$ and $x - 2y + 5 = 0$
Here, $a_1 = 2, b_1 = 1, c_1 = 3$ and $a_2 = 1, b_2 = -2, c_2 = 0$
$\therefore\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=2\times1+1\times(-2)+3\times0=0$
Thus, the given planes are perpendicular to each other.
View full question & answer→Question 1003 Marks
Show that the line through points $(4, 7, 8)$ and $(2, 3, 4)$ is parallel to the line throught the points $(-1, -2, 1)$ and $(1, 2, 5).$
AnswerSuppose the points are $A(2,3,4), B(-1,-2,1)$ and $C(5,8,7)$.
We know that the direction ratios of the line passing through the points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_2\right)$ are $x_2-x_1, y_2-y_1, z_2$ - $z _1$.
Let the first two points be $A(4,7,8)$ and $B(2,3,4)$.
Thus, the direction ratios of $A B$ are $(2-4),(3-7),(4-8)$, i.e. $-2,-4,-4$.
Similarly, Let the other two points be $C(-1,-2,1)$ and $D(1,2,5)$.
Thus, the direction ratios of $C D$ are $[1-(-1)],[2-(-2)],(5-1)$, i.e. $2,4,4$.
It can be seen that the direction ratios of $C D$ are -1 times that of $A B$, i.e. they are proportional. Therefore, $A B$ and $C D$ are parallel lines.
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