Question 1514 Marks
Show that the four points (0, -1, -1), (4, 5, 1), (3, 9, 4) and (-4, 4, 4) are coplanar and find the equation of the common plane.
AnswerThe equation of the plane passing through points (0, -1, -1), (4, 5, 1) and (3, 9, 4) is given by,
$\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}+1\\4-0&5+1&1+1\\3-0&9+1&4+1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}&\text{y}+1&\text{z}+1\\4&6&2\\3&10&5\end{vmatrix}=0$
$\Rightarrow10\text{x}-14(\text{y}+1)+22(\text{z}+1)=0$
$\Rightarrow5\text{x}-7(\text{y}+1)+11(\text{z}+1)=0$
$\Rightarrow5\text{x}-7\text{y}+11\text{z}+4=0$
Substituting the last points (-4, 4, 4) (it means x = -4; y = 4; z = 4) in this plane equation, we get
5(-4) - 7(4) + 11(4) + 4 = 0
⇒ -48 + 48 = 0
So, the plane equation is satisfied by the points (-4, 4, 4)
So, the given pointsa are coplanar and the equation of the common plane (as we already founded) is
5x - 7y + 11z + 4 = 0
View full question & answer→Question 1524 Marks
Find the vector and Cartesian equations of the line through the point (1, 2, – 4) and perpendicular to the two lines.
$\overrightarrow{\text{r}} = (8\hat{\text{i}} - 19\hat{\text{j}} + 10\hat{\text{k}})+\lambda(3\hat{\text{i}} - 16\hat{\text{j}} + 7\hat{\text{k})}$ and $\overrightarrow{\text{r}} = (15\hat{\text{i}} - 29\hat{\text{j}} + 5\hat{\text{k}})+\mu(3\hat{\text{i}} - 8\hat{\text{j}} + 5\hat{\text{k})}.$
AnswerVector equation of the required line is
$\overrightarrow{\text{r}} = (\hat{\text{i}} + 2\hat{\text{j}} - 4\hat{\text{k}})+\mu[(3\hat{\text{i}} - 16\hat{\text{j}} + 7\hat{\text{k})}\times(3\hat{\text{i}} + 8\hat{\text{j}} - 5\hat{\text{k})]}$
$\Rightarrow\overrightarrow{\text{r}} = (\hat{\text{i}} + 2\hat{\text{j}} - 4\hat{\text{k}})+\lambda[(2\hat{\text{i}} + 3\hat{\text{j}} + 6\hat{\text{k})]}$
in cartesian form, $\frac{\text{x - 1}}{2} = \frac{\text{y - 2}}{3} = \frac{\text{z + 4}}{6}$
View full question & answer→Question 1534 Marks
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.
AnswerThe equation of plane through the intersection of planes, x + y + z = 1 and 2x + 3y + 4z = 5, is
$(\text{x + y + z}-1)+\lambda(2\text{x}+3\text{y}+4\text{z}-5)=0$
$\Rightarrow\ \ (2\lambda+1)\text{x}+(3\lambda+1)\text{y}+(4\lambda+1)\text{z}-(5\lambda+1)=0\ \ ....(1)$
The direction ratios, a1, b1, c1, are 1, -1, and 1.
Since the planes are perpendicular,
a1a2 + b1b2 + c1c2 = 0
$\Rightarrow\ \ (2\lambda+1)-(3\lambda+1)+(4\lambda+1)=0$
$\Rightarrow\ \ 3\lambda+1=0$
$\Rightarrow\ \ \lambda=-\frac{1}{3}$
Substituting $\lambda=-\frac{1}{3}$ in equation (1), we obtain
$\frac{1}{3}\text{x}-\frac{1}{3}\text{z}+\frac{2}{3}=0$
⇒ x - z + 2 = 0
This is the required equation of the plane.
View full question & answer→Question 1544 Marks
Find the value of $\lambda$ such that the line $\frac{\text{x}-2}{6}=\frac{\text{y}-1}{\lambda}=\frac{\text{z}+5}{-4}$ is perpendicular to the plane 3x - y - 2z = 7.
AnswerHere, given mid line $\frac{\text{x}-2}{6}=\frac{\text{y}-1}{\lambda}=\frac{\text{z}+5}{-4}$ is parpendicular to plane 3x - y - 2z = 7 so, normal vector of plane is parallel to line so,
Direction ratios of normal to plane are proparional to the direction ratios of line.
Here,
$\frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}$
cross multiplying the last two
$-2\lambda=4$
$\lambda=\frac{4}{-2}$
$\lambda=-2$
View full question & answer→Question 1554 Marks
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: $\frac{\text{x}-8}{3}=\frac{\text{y}+19}{-16}=\frac{\text{z}-10}{7}\ \text{and}\ \frac{\text{x}-15}{3}=\frac{\text{y}-29}{8}=\frac{\text{z}-5}{-5}.$
AnswerGiven: A point on the Required line is A(1, 2, -4)
$\therefore\ \text{Position vector of point A is }\vec{\text{a}}=(1,2,-4)=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
Also given equations of two lines
$\frac{\text{x}-8}{3}=\frac{\text{y}+19}{-16}=\frac{\text{z}-10}{7}\ \text{and}\ \frac{\text{x}-15}{3}=\frac{\text{y}-29}{8}=\frac{\text{z}-5}{-5}$
$\therefore$ Direction ratios of given two lines are
$\vec{\text{b}_1}=3\hat{\text{i}}-16\hat{\text{j}}+7\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}$
$\vec{\text{b}}=\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-16&7\\3&8&-5\end{vmatrix}$
Expanding along first row,
$=\hat{\text{i}}(80-56)-\hat{\text{j}}(-15-21)+\hat{\text{k}}(24+48)=24\hat{\text{i}}+36\hat{\text{j}}+72\hat{\text{k}}$
$\Rightarrow\ \vec{\text{b}}=12\Big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\Big)$
$\therefore$ Equation of the required line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow \vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\Big)+\lambda(12)\Big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\Big)$
Again replacing $12\lambda\ \text{by}\ \lambda$
$\Rightarrow \vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\Big)+\lambda\Big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\Big).$
View full question & answer→Question 1564 Marks
Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.
AnswerThe vectors, represented by these are
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\Big|\vec{\text{a}}\Big|\Big|\vec{\text{b}}\Big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{(2)^2+(3)^2+(6)^2}\sqrt{(1)^2+(2)^2+(2)^2}}$
$=\frac{(2)(1)+(3)(2)+(6)(2)}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{2+6+12}{\sqrt{49}\sqrt{9}}$
$=\frac{20}{7\times3}$
$\cos\theta=\frac{20}{21}$
$\theta=\cos^{-1}\Big(\frac{20}{21}\Big)$
Angle between the lines $=\cos^{-1}\Big(\frac{20}{21}\Big)$.
View full question & answer→Question 1574 Marks
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane.
AnswerGiven: A line through the points A(5, 1, 6) and B(3, 4, 1)
$\therefore$ Direction ratios of this line AB are x2 - x1, y2 - y1, z2 - z1
⇒ 3 - 5, 4 - 1, 1 - 6
⇒ -2, 3, -5 = a, b, c
Equation of the line AB is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$\Rightarrow\ \ \frac{\text{x}-5}{-2}=\frac{\text{y}-1}{3}=\frac{\text{z}-6}{-5}\ \ \ .....(\text{i})$
Now we have to find the coordinates of the point where this line AB crosses the YZ-plane
i.e., x = 0 .......(ii)
Putting x = 0 in eq. (i), we get
$\frac{-5}{-2}=\frac{\text{y}-1}{3}=\frac{\text{z}-6}{-5}$
$\Rightarrow\ \ \frac{\text{y}-1}{3}=\frac{5}{2}\ \text{and}\ \frac{\text{z}-6}{-5}=\frac{5}{2}$
⇒ 2y - 2 = 15 and 2z - 12 = -25
⇒ 2y = 17 and 2z = -13
$\Rightarrow\ \ \ \text{y}=\frac{17}{2}\ \text{and}\ \text{z}=\frac{-13}{2}$
Thus, required point is $\text{P}\Big(0,\ \frac{17}{2},\ \frac{-13}{2}\Big).$
View full question & answer→Question 1584 Marks
Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.
AnswerEquation of line through A(3, 4, 1) and B(5, 1, 6)
$\frac{\text{x - 3}}{2} = \frac{\text{y - 4}}{-3} = \frac{\text{z - 1}}{5} = \text{k (say)}$
General point on the line:
$\text{x = 2k + 3, y = -3k + 4, z = 5k + 1}$
line crosses xz plane i.e. $\text{y = 0 if -3k + 4 = 0}$
$\therefore\text{k} = \frac{4}{3}$
Co-ordinate of required point $\bigg(\frac{17}{3}, 0, \frac{23}{3}\bigg)$
Angle, which line makes with xz plane:
$\sin\theta = \bigg|\frac{2(0) + (-3) (1) + 5 (0)}{\sqrt{4 + 9 + 25}\sqrt{1}}\bigg| = \frac{3}{\sqrt{38}}\Rightarrow \theta = \sin^{-1}\bigg(\frac{3}{\sqrt{38}}\bigg)$
View full question & answer→Question 1594 Marks
Find the vector equation of the plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3).
Answer
The required plane passes through the point P(2, 5, -3), whose position vector is $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}}$ and is normal to the vector $\vec{\text{n}}$ given by
$\overrightarrow{\text{PQ}}=\overrightarrow{\text{OQ}}-\overrightarrow{\text{OP}}$
$=(-2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})-(2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}})$
$=-4\hat{\text{i}}-8\hat{\text{j}}+8\hat{\text{k}}$
$\overrightarrow{\text{PR}}=\overrightarrow{\text{OR}}-\overrightarrow{\text{OP}}$
$=(5\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})-(2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}})$
$=3\hat{\text{i}}-2\hat{\text{j}}-0\hat{\text{k}}$
$\vec{\text{n}}=\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-4&-8&8\\3&-2&0\end{vmatrix}$
$=16\hat{\text{i}}+24\hat{\text{j}}+32\hat{\text{k}}$
The vector equation of the required plane is
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(16\hat{\text{i}}+24\hat{\text{j}}+32\hat{\text{k}})=(2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}})\cdot(16\hat{\text{i}}+24\hat{\text{j}}+32\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\Big[8(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\Big]=32+120-96$
$\Rightarrow\vec{\text{r}}\cdot\Big[8(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\Big]=56$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})=7$
View full question & answer→Question 1604 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 1614 Marks
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).
AnswerDirection ratios of the line joining A and B are -1 - 3, 1 - 5, 2 - (-4)
$\Rightarrow-4,\ -4,\ 6\ \ \ \ \ \ \ \ [\because\ \text{x}_2-\text{x}_1,\ \text{y}_2-\text{y}_1]$
$\therefore$ Direction cosines of line AB are
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
$\Rightarrow\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}}$
$\Rightarrow\ \frac{-4}{\sqrt{16+16+36}},\ \frac{-4}{\sqrt{16+16+36}},\ \frac{6}{\sqrt{16+16+36}}$
$\Rightarrow\ \frac{-4}{\sqrt{68}},\ \frac{-4}{\sqrt{68}},\ \frac{6}{\sqrt{68}}\ \ \Rightarrow\ \frac{-4}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}},\ \frac{6}{2\sqrt{17}}$
$\Rightarrow\ \frac{-2}{\sqrt{17}},\ \frac{-2}{\sqrt{17}},\ \frac{3}{\sqrt{17}}$
Now Direction ratios of the line joining B and C are -5 - (-1), -5 - 1, -2 - 2 = -4, -6, -4
$\therefore$ Direction cosines of line BC are
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
$\therefore$ Direction cosines of line BC are
$\Rightarrow\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}$
$\Rightarrow\ \frac{-4}{\sqrt{16+36+16}},\ \frac{-6}{\sqrt{16+36+16}},\ \frac{-4}{\sqrt{16+36+16}}$
$\Rightarrow\ \frac{-4}{\sqrt{68}},\ \frac{-6}{\sqrt{68}},\ \frac{-4}{\sqrt{68}}\ \ \Rightarrow\ \frac{-4}{2\sqrt{17}},\ \frac{-6}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}}$
$\Rightarrow\ \frac{-2}{\sqrt{17}},\ \frac{-3}{\sqrt{17}},\ \frac{-2}{\sqrt{17}}$
Direction ratios of the line joining C and A are 3 - (-5), 5 - (-5), -4 - (-2) = 8, 10, -2
$\therefore$ Direction cosines of line CA are
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
$\Rightarrow\ \frac{8}{\sqrt{(8)^2+(10)^2+(-2)^2}},\ \frac{10}{\sqrt{(8)^2+(10)^2+(-2)^2}},\ \frac{-2}{\sqrt{(8)^2+(10)^2+(-2)^2}}$
$\Rightarrow\ \frac{8}{\sqrt{64+100+4}},\ \frac{10}{\sqrt{64+100+4}},\ \frac{-2}{\sqrt{64+100+4}}$
$\Rightarrow\ \frac{8}{\sqrt{168}},\ \frac{10}{\sqrt{168}},\ \frac{-2}{\sqrt{168}}\ \ \Rightarrow\ \frac{8}{2\sqrt{42}},\ \frac{10}{2\sqrt{42}},\ \frac{-2}{2\sqrt{42}}$
$\Rightarrow\ \frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ \frac{-1}{\sqrt{42}}.$
View full question & answer→Question 1624 Marks
Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.
AnswerEquation of line through A(3, 4, 1) and B(5, 1, 6)
$\frac{\text{x - 3}}{2} = \frac{\text{y - 4}}{-3} = \frac{\text{z - 1}}{5} = \text{k (say)}$
General point on the line:
$\text{x = 2k + 3, y = -3k + 4, z = 5k + 1}$
line crosses xz plane i.e. $\text{y = 0 if -3k + 4 = 0}$
$\therefore\text{k} = \frac{4}{3}$
Co-ordinate of required point $\bigg(\frac{17}{3}, 0, \frac{23}{3}\bigg)$
Angle, which line makes with xz plane:
$\sin\theta = \bigg|\frac{2(0) + (-3) (1) + 5 (0)}{\sqrt{4 + 9 + 25}\sqrt{1}}\bigg| = \frac{3}{\sqrt{38}}\Rightarrow \theta = \sin^{-1}\bigg(\frac{3}{\sqrt{38}}\bigg)$
View full question & answer→Question 1634 Marks
Find the value of p, so that the lines $l_1$ :$\frac{1-\text{x}}{3}=\frac{7\text{y}-\text{14}}{\text{p}}=\frac{\text{z}-\text{3}}{2}$ and $l_2$: $\frac{7-\text{7x}}{3\text{p}}=\frac{\text{y}-\text{5}}{1}=\frac{6-\text{z}}{5}$are perpendicular to each other. Also find the equations of a line passing through a point (3, 2,– 4) and parallel to line $l_1$.
AnswerGiven lines can be written as
$l_1$:$\frac{1-\text{x}}{-3}=\frac{\text{y}-\text{2}}{\text{p}/7}=\frac{\text{z}-\text{3}}{2}$ ; $l_2$: $\frac{\text{x}-\text{1}}{-3\text{p}/7}=\frac{\text{y}-\text{5}}{1}=\frac{\text{z}-6}{-5}$
since the lines are perpendicular
$\therefore\ (-3)\big(-\frac{3\text{p}}{7}\big)+\big(\frac{\text{p}}{7}\big)(1)+(2)(-5)=0$
$\Rightarrow$ p = 7
Equation of line passing through (3, 2, – 4) and parallel to $l_1$ is
$\frac{\text{x}-3}{-3}=\frac{\text{y}-\text{2}}{\text{1}}=\frac{\text{z}+\text{4}}{2}$
View full question & answer→Question 1644 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.
View full question & answer→Question 1654 Marks
Find the vector and the cartesian equations of the lines that passes through the origin and (5, -2, 3).
Answer$\vec{\text{a}}=$ Position vector of a point here O (say) on the line $(0,\ 0,\ 0)=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\vec{0}$ $\vec{\text{b}}=$ A vector along the line $=\overrightarrow{\text{OA}}=$ Position vector of a point A - Position vector of O $=(5,-2,\ 3)-(0,\ 0,\ 0)=(5,-2,\ 3)=5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ $\therefore$ Vector equation of the line is $\Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)$ $\Rightarrow\ \ \vec{\text{r}}=\vec{0}+\lambda\Big(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)$ $\Rightarrow\ \ \vec{\text{r}}=\lambda\Big(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)$ Now Cartesian equation of the line Direction ratios of line OA are 5 - 0, -2 - 0, 3 - 0 = 5, -2, 3 And a point on the line is O(0, 0, 0) = (x1, y1, z1) $\therefore$ Cartesian equation of the line $=\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$ $=\frac{\text{x}-0}{5}=\frac{\text{y}-0}{-2}=\frac{\text{z}-0}{3}=\frac{\text{x}}{5}=\frac{\text{y}}{-2}=\frac{\text{z}}{3}$ Remark: In the solution of the above question we can also take:
$\vec{\text{a}}=$ Position vector of point A = (5, -2, 3) $=5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ for vector form and point A as (x1, y1, z1) = (5, -2, 3) for Cartesian form.
Then the equation of the line in vector form is $\vec{\text{r}}=5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}+\lambda\Big(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)$
And equation of line in Cartesian form is $\frac{\text{x}-5}{5}=\frac{\text{y}+2}{-2}=\frac{\text{z}-3}{3}$
View full question & answer→Question 1664 Marks
Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
AnswerSince equation of any plane through the point (-1, 3, 2) is a(x - x1) + b(y - y1) + c(z - z1) = 0
$\therefore$ a(x + 1) + b(y - 3) + c(z - 2) .....(i)
⇒ ax + a + by - 3b + cz - 2c = 0
⇒ ax + by + cz = -a + 3b + 2c
This required plane is perpendicular to the olane x + 2y + 3z = 5(a1a2 + b1b2 + c1c2 = 0)
$\therefore$ Products of coefficients $\Rightarrow\ \text{a}(1)+\text{b}(2)+\text{c}(3)=0\ \ \ ....(\text{ii})$
Again, the required plane is perpendicular to the plane 3x + 3y + z = 0
$\therefore$ Products of coefficients $\Rightarrow\ \text{a}(3)+\text{b}(3)+\text{c}(1)=0\ \ \ ....(\text{iii})$
Solving eq. (ii) and (iii), we get
$\frac{\text{a}}{2-9}=\frac{\text{b}}{9-1}=\frac{\text{c}}{3-6}\ \ \ \Rightarrow\ \ \frac{\text{a}}{-7}=\frac{\text{b}}{8}=\frac{\text{c}}{-3}$
Putting these values of a, b, c in eq. (i), we get
-7(x + 1) + 8(y - 3) - 3(z - 2) = 0
⇒ -7x - 7 + 8y - 24 - 3z + 6 = 0
⇒ -7x + 8y - 3z - 25 = 0
⇒ 7x - 8y + 3z + 25 = 0.
View full question & answer→Question 1674 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 1684 Marks
Find the value of $\lambda$ for which the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{\lambda^2}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{\lambda^2}=\frac{\text{z}-1}{2}$ are coplanar.
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 $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{\lambda^2}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{\lambda^2}=\frac{\text{z}-1}{2}$ are coplanar.
$\therefore\ \begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}3-1&2-2&1-(-3)\\1&2&\lambda^2\\1&\lambda^2&2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}2&0&4\\1&2&\lambda^2\\1&\lambda^2&2\end{vmatrix}=0$
$\Rightarrow2(4-\lambda^4)-0+4(\lambda^4-2)=0$
$\Rightarrow-2\lambda^4+4\lambda^2=0$
$\Rightarrow\lambda^2(\lambda^2-2)=0$
$\Rightarrow\lambda^2=0\text{ or }\lambda^2-2=0$
$\Rightarrow\lambda=0\text{ or }\lambda=\pm\sqrt{2}$
Thus, the values of $\lambda$ are $0,-\sqrt{2}$ and $\sqrt{2}$
View full question & answer→Question 1694 Marks
Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XY-plane.
AnswerEquations of line AB are
$\frac{\text{x-3}}{\text{2}}=\frac{\text{y-4}}{-3}=\frac{\text{z-1}}{5}=\lambda\ ......\text{(i)}$
the general point on (i) is
$2\lambda + 3,\ – 3\lambda + 4,\ 5\lambda + 1$
the line (i) crosses XY-plane, then z = 0
$\lambda=-\frac{1}{5}$
hence point is $\bigg[2\Big(\frac{-1}{5}\Big)+3,\ -3\Big(\frac{-1}{5}\Big)+4,\ 5\Big(\frac{-1}{5}\Big)+1\bigg]$
$=\Big[\frac{13}{5},\ \frac{23}{5},\ 0\Big]$
View full question & answer→Question 1704 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$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$
$\vec{\text{a}}_1=4\hat{\text{i}}-\hat{\text{j}}\ \ \ \vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big)$
$\vec{\text{a}}_2=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}_2=2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\cdot\big(\vec{\text{b}}_2\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1-\vec{\text{b}}_2\big|}\end{vmatrix}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}} &\hat{\text{j}}&\hat{\text{k}}\\1 & 2&-3\\2&4&-5 \end{vmatrix}$
$=\hat{\text{i}}(-10+12)-\hat{\text{j}}(-5+6)+\hat{\text{k}}(4-4)$
$=2\hat{\text{i}}-\hat{\text{j}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(-3\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)\cdot\big(2\hat{\text{i}}-\hat{\text{j}}\big)}{\sqrt{4+1}}\end{vmatrix}$
$=\begin{vmatrix}\frac{-6}{\sqrt{5}}\end{vmatrix}=\frac{6}{\sqrt{5}}$
View full question & answer→Question 1714 Marks
$\overrightarrow{\text{n}}$ is a vector of magnitude $\sqrt{3}$ and is equally inclined to an acute angle with the coordinate axes. Find the vector and cartesian form of the equation of a plane which passes through (2, 1, -1) and is normal to $\overrightarrow{\text{n}}$
AnswerHere, it is given that $\vec{\text{n}}=\sqrt{3}$ and $\vec{\text{n}}$ makes equal angle with coordinate axes.
Let, $\vec{\text{n}}$ has direction cosine as l. m and n and it makes angle of $\alpha,\beta$ and $\gamma$ with the coordinate axes, so
Here, $\alpha=\beta=\gamma$
$\Rightarrow\cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\text{l}=\text{m}=\text{n}=\text{p}(\text{say})$
We know that,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\text{p}^2+\text{p}^2+\text{p}^2=1$
$3\text{p}^2=1$
$\text{p}^2=\frac{1}{3}$
$\text{p}=\pm\frac{1}{\sqrt{3}}$
So,
$\text{l}=\pm\frac{1}{\sqrt{3}}$
$\cos\alpha=\pm\frac{1}{\sqrt{3}}$
Now, $\alpha=\cos^{-1}\Big(-\frac{1}{\sqrt{3}}\Big)$
It gives, $\alpha$ is an obtuse angle so, neglect it.
Again, $\alpha=\cos^{-1}\Big(-\frac{1}{\sqrt{3}}\Big)$
It gives, $\alpha$ is an acute angle, so
$\cos\alpha=\frac{1}{\sqrt{3}}$
$\therefore\ \text{l}=\text{m}=\text{n}=\frac{1}{\sqrt{3}}$
So,
$\vec{\text{n}}=|\vec{\text{n}}|(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}})$
$=\sqrt{3}\Big(\frac{1}{\sqrt{3}}\hat{\text{i}}+\frac{1}{\sqrt{3}}\hat{\text{j}}+\frac{1}{\sqrt{3}}\hat{\text{k}}\Big)$
$\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
And, $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
We know that, vector equation of a plane passing through the point $\vec{\text{a}}$ and perpendicular to the vector $\vec{\text{n}}$ is given by,
$(\vec{\text{r}}-\vec{\text{a}}){\vec{\text{n}}}=0$
$\big[\vec{\text{r}}-(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})\big]\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})-(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})-[(2)(1)+(1)(1)+(-1)(1)]=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})-[2+1-1]=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})-2=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2$
Put, $\vec{\text{r}}=(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})$
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2$
$(\text{x})(1)+(\text{y})(1)+(\text{z})(1)=2$
$\text{x}+\text{y}+\text{z}=2$
So, vector and cartesian equation of the plane is,
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2$
$\text{x}+\text{y}+\text{z}=2$
View full question & answer→Question 1724 Marks
Find the equation of the plane through the points (2, 1, -1) and (-1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
AnswerThe equation of the plane passing through (2, 1, -1) is
a(x - 2) + b(y - 1) + c(z + 1) = 0 ......(i)
Since, this passes through (-1, 3, 4).
$\therefore$ a(-1 - 2) + b(3 - 1) + c(4 + 1) = 0
⇒ -3a + 2b + 5c = 0 ......(ii)
Since, the plane (i) is perpendicular to the plane x - 2y + 4z = 10.
$\therefore$ 1× a - 2 × b + 4 × c = 0
⇒ a - 2b + 4c = 0 ......(iii)
On solving equations (ii) and (iii), by cross multiplication method, we get
$\frac{\text{a}}{8+10}=\frac{\text{-b}}{-17}=\frac{\text{c}}{4}=\lambda$
$\Rightarrow\text{a}=18\lambda,\text{b}=17\lambda,\text{c}=4\lambda$
From Eq. (i),
$18\lambda(\text{x}-2)+17\lambda(\text{y}-1)+4\lambda(\text{z}-1)=0$
$\Rightarrow18\text{x}-36+17\text{y}-17-4\text{z}+4=0$
$\therefore18\text{x}+17\text{y}+4\text{z}-49$
View full question & answer→Question 1734 Marks
Find the equation of the perpendicular drawn from the point P(-1, 3, 2) to the line $\vec{\text{r}}=\big(2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big).$ Also, find the coordinates of the foot of the perpendicular from P.
AnswerLet Q be the perpendicular drow from $\text{p}\big(\hat{-\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)$ on the line
$\vec{\text{r}}=\big(2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)$
Let the position vector of Q be
$\big (2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)$
$\big(2\lambda\big)\hat{\text{i}}+\big(2+\lambda\big)\hat{\text{j}}+\big(3+3\lambda\big)\hat{\text{k}}$
$\overrightarrow{\text{PQ}}=$ position vector of Q-position vector of P
$\big\{\big(2\lambda\big)\hat{\text{i}}+\big(2+\lambda\big)\hat{\text{j}}+\big(3+3\lambda\big)\hat{\text{k}}\big\}-\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)$
$=(2\lambda+1)\hat{\text{i}}+(2\lambda-3)\hat{\text{j}}+(3+3\lambda-2)\hat{\text{k}}$
$\overrightarrow{\text{PQ}}=(2\lambda+1)\hat{\text{i}}+(\lambda-1)\hat{\text{j}}+(3\lambda+1)\hat{\text{k}}$
Since, $\overrightarrow{\text{PQ}}$ is perpendicular to given line, so
$\big\{(2\lambda+1)\hat{\text{i}}+(\lambda-1)\hat{\text{j}}+(3\lambda+1)\hat{\text{k}}\big\}\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$(2\lambda+1)(2)+(\lambda-1)(1)+(3\lambda+1)3=0$
$4\lambda+2+\lambda-1+9\lambda+3=0$
$14\lambda+4=0$
$\lambda=-\frac{4}{14}$
$\lambda=-\frac{2}{7}$
Position vector of Q $=(2\lambda)\hat{\text{i}}+(2+\lambda)\hat{\text{j}}+(3+3\lambda)\hat{\text{k}}$
$=2\Big(-\frac{2}{7}\Big)\hat{\text{i}}+\Big(2-\frac{2}{7}\Big)\hat{\text{j}}+\Big(3+3\Big(-\frac{2}{7}\Big)\Big)\hat{\text{k}}$
$=-\frac{4}{7}\hat{\text{i}}+\frac{12}{7}\hat{\text{j}}+\frac{15}{7}\hat{\text{k}}$
Coordinates of foot of the perpendicular $=\Big(-\frac{4}{7},\frac{12}{7},\frac{15}{7}\Big)$
Equation of PQ is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\big(\vec{\text{b}}-\vec{\text{a}}\big)$
$\Rightarrow\vec{\text{r}}=\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)+\lambda\Big(\Big(-\frac{4}{7}\hat{\text{i}}+\frac{12}{7}\hat{\text{j}}+\frac{15}{7}\hat{\text{k}}\Big)-\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)\Big)$
View full question & answer→Question 1744 Marks
Find the shortest distance between the lines whose vector equations are:
$\vec{\text{r}}=(1-\text{t})\hat{\text{i}}+(\text{t}-2)\hat{\text{j}}+(3-2\text{t})\hat{\text{k}}\ \text{and}$
$\vec{\text{r}}=(\text{s}+1)\hat{\text{i}}+(2\text{s}-1)\hat{\text{j}}-(2\text{s}+1)\hat{\text{k}}$
AnswerEquation of first line is $\vec{\text{r}}=(1-\text{t})\hat{\text{i}}+(\text{t}-2)\hat{\text{j}}+(3-2\text{t})\hat{\text{k}}$
$\hat{\text{i}}-\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}-2\hat{\text{j}}+3\hat{\text{k}}-2\text{t}\hat{\text{k}}$
$=\Big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)+\text{t}\Big(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{a}_1}+\text{t}\vec{\text{b}_1},$
$\vec{\text{a}_1}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\ \ \ \vec{\text{b}_1}=-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
Equation of second line is $\vec{\text{r}}=(\text{s}+1)\hat{\text{i}}+(2\text{s}-1)\hat{\text{j}}+(2\text{s}+1)\hat{\text{k}}$
$\text{s}\hat{\text{i}}+\hat{\text{i}}+2\text{s}\hat{\text{j}}-\hat{\text{j}}-2\text{s}\hat{\text{k}}-\hat{\text{k}}$
$=\Big(\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\Big)+\text{s}\Big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{a}_2}+\text{s}\vec{\text{b}_2},$
$\vec{\text{a}_2}=\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}},\ \ \ \vec{\text{b}_2}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
Now shortest distance $(\text{d})=\frac{\Big|\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)\Big|}{\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|}\ \ \ \ ...(\text{i})$
$\vec{\text{a}_2}-\vec{\text{a}_1}=\Big(\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\Big)-\Big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)=\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-1&1&-2\\1&2&-2\end{vmatrix}$
$=(-2+4)\hat{\text{i}}-(2+2)\hat{\text{j}}+(-2-1)\hat{\text{k}}=2\hat{\text{i}}-4\hat{\text{j}}-3\hat{\text{k}}$
$\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|=\sqrt{(2)^2+(-4)^2+(-3)^2}=\sqrt{29}$
$\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)=\Big(\hat{\text{j}}-4\hat{\text{k}}\Big).\Big(2\hat{\text{i}}-4\hat{\text{j}}-3\hat{\text{k}}\Big)$
=0 × 2 + 1 × (-4) + (-4)(-3) = 8
Putting these values in eq.(i),
Shortest distance $(\text{d})=\frac{|8|}{\sqrt{29}}=\frac{8}{\sqrt{29}}.$
View full question & answer→Question 1754 Marks
Find the shortest distance between the following pairs of lines whose cartesian equation are:
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\text{z}$ and $\frac{\text{x}+2}{3}=\frac{\text{y}-2}{1};\text{z}=2$
AnswerThe equation of the given are
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\frac{\text{z}-0}{1}\dots(1)$
$\frac{\text{x}+1}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-2}{0}\dots(2)$
Since line (1) passes through the point (1, -1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$
Here,
$\vec{\text{a}}_1=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{b}}_1=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
Also, line (2) passes through the point (-1, 2, 2) and has diraction ratios proportional to 3, 1, 0.
Its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$
Here,
$\vec{\text{a}}_2=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=-2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&1\\3&1&0\end{vmatrix}$
$=-\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-1)^2+3^2(-7)^2}$
$=\sqrt{1+9+49}$
$=\sqrt{59}$
and $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big).\big(-\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}\big)$
$=2+9-14$
$=-3$
The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{-3}{\sqrt{59}}\Big|$
$=\frac{3}{\sqrt{59}}$
View full question & answer→Question 1764 Marks
Find the distance between the point (-1, -5, -10) and the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5.$
AnswerAny point on the line $\frac{\text{x} - 2}{3} = \frac{\text{y} + 1}{4} = \frac{\text{z} - 2}{12} = \text{is} (3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$
If this is the point of intersection with plane $\text{x - y + z = 5}$
$\text{then 3} \lambda + 2 - 4\lambda + 1 + 12\lambda + 2 - 5 = 0 \Rightarrow \lambda = 0$
$\therefore$ Point of intersection is (2, -1, 2)
Required distance = $\sqrt{(2 + 1)^{2} +(-1 + 5)^{2} + (2 + 10)^{2} } = 13$
View full question & answer→Question 1774 Marks
Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4).
AnswerEquation of line $\vec{\text{AB}}$
$\vec{\text{r}} = (-\hat{\text{j}} - \hat{\text{k}}) + \lambda (4\hat{\text{i}} + 6\hat{\text{j}} + 2\hat{\text{k}})$
Equation of line $\vec{\text{CD}}$
$\vec{\text{r}} = (3\hat{\text{i}} + 9\hat{\text{j}} + 4\hat{\text{k}}) + \mu (-7\hat{\text{i}} - 5\hat{\text{j}})$
$\vec{\text{a}}_{2} - \vec{\text{a}}_{1} = 3\hat{\text{i}} + 10\hat{\text{j}} + 5 \hat{\text{k}}$
$\vec{\text{b}}_{1} \times \vec{\text{b}}_{2} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{vmatrix} = 10\hat{\text{i}} - 14\hat{\text{j}} + 22\hat{\text{k}}$
$(\vec{\text{a}}_{2} - \vec{\text{a}}_{1}). (\vec{\text{b}}_{1} \times \vec{\text{b}}_{2}) = 30 – 140 + 110 = 0$
$\Rightarrow$ Lines intersect.
View full question & answer→Question 1784 Marks
Show that the lines $\frac{5 - x }{-4} =\frac{\text{y}- 7}{4} = \frac{\text{z} + 3}{-5}\text{ and } \frac{{x} - 8}{7} =\frac{2\text{y} - 8}{2} =\frac{\text{z} - 5 }{3}$ are coplanar.
AnswerEquations of lines are:
$\frac{x - 5}{4} =\frac{\text{y}- 7}{4} = \frac{\text{z} + 3}{-5};\ \frac{{x} - 8}{7} =\frac{\text{y} - 4}{1} =\frac{\text{z} - 5 }{3}$
Here, x1 = 5, y1 = 7, z1 = – 3 ; x2 = 8, y2 = 4, z2 = 5
a1 = 4, b1 = 4, c1 = – 5 ; a2 = 7, b2 = 1, c2 = 3
$ \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 & -3 & 8 \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{vmatrix}= 3(17) + 3 (47) + 8 (– 24) = 0$
$\therefore$ lines are co-planar
View full question & answer→Question 1794 Marks
Find the equation of the plane which is parallel to 2x - 3y + z = 0 and which passes through (1, -1, 2).
AnswerGiven, equation of plane is,
2x - 3y + z = 0 ....(i)
We know that equation of a plane parallel the plane (i) is given by
$2\text{x}-3\text{y}+\text{z}+\lambda=0\ ....(\text{ii})$
Given that, plane (ii) is passing through the point (1, -1, 2) so it must satisfy the equation (ii),
$2(1)-3(-1)+(2)+\lambda=0$
$2+3+2+\lambda=0$
$7+\lambda=0$
$\lambda=-7$
Put the value of $\lambda$ in equation (ii),
2x - 3y + z - 7 = 0
So, equation of the required plane is,
2x - 3y + z = 7
View full question & answer→Question 1804 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 1814 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 l2 + m2.
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) Ax1 + By1 + Cz1 + 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 × 3 + m × (-2) - (-4) = 9
⇒ 3l - 2m = 5 ....(i)
Also,
2 × l + (-1) × m + 3 × (-1) = 0
⇒ 2l - m = 3 ....(ii)
Solving (i) and (ii) we get
l = 1 and m = -1
$\therefore$ l2 + m2 = 12 + (-1)2 = 1 + 1 = 2
Thus, the value of l2 + m2 is 2
View full question & answer→Question 1824 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.
View full question & answer→Question 1834 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 1844 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 1854 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 1864 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 1874 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 c1 = 1 and c2 = 2, find c3 which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
- If c2 = –1 and c3 = 1, show that no value of c1 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 1884 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 1894 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 1904 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 1914 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 1924 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 1934 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 1944 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 a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 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 1954 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
x1 = -1, y1 = -1, z1 = -1,
a1 = 7, b1 = -6, c1 = 1
Again equation of another line is $\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}$
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
x2 = 3, y2 = 5, z2 = 7,
a2 = 1, b2 = -2, c2 = 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 1964 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 1974 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 1984 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 1994 Marks
Find the distance of the point (3, 3, 3) from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$
AnswerThe given plane is
$\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$
$\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=9$
We know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ is given by
$\text{p}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
Finding the distance from (3, 3, 3) $($which means $3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})$ to the given plane
Here, $\vec{\text{a}}=3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}},\vec{\text{n}}=5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}},\text{d}=9$
So, the required distance p
$=\frac{\big|(3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})-9\big|}{\big|3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}\big|}$
$=\frac{|-15-6+21-9|}{\sqrt{25+4+49}}$
$=\frac{-9}{\sqrt{78}}$
$=\frac{-9}{\sqrt{78}}\text{ units}$
View full question & answer→Question 2004 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 $=\frac{2}{3},\frac{1}{3},-\frac{2}{3}$
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