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The plane · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 28 questions.

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Question 12 Marks
Find the vector equation of a plane which is at a distance of 3 units from the origin and has $\hat{\text{k}}$ as the unit vector normal to it.
Answer
Here, it is given that, the required plane is at a distance of 3 unit from origin and k is unit vector normal to it. we know that, vector equation of a plane normal to unit vector $\hat{\text{n}}$ and at distance d from origin, is
$\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$
So, here d = 3 units
$\hat{\text{n}}=\hat{\text{k}}$
The equation of the required plane is,
$\vec{\text{r}}\cdot\hat{\text{k}}=3$
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Question 22 Marks
Find the vector equation one of following plane.
x + y - z = 5
Answer
Given, equation of plane is,
x + y - z = 5
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})=5$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=5$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=5$
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Question 32 Marks
Find the vector equation one of following plane.
2x - y + 2z = 8
Answer
Given, equation of plane is,
2x - y + 2z = 8
 $(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=8$
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Question 42 Marks
Write the equation of the plane parallel to XOY- plane and passing through the point (2, -3, 5).
Answer
The equation of the plane parallel to the plane XOY is z = b .....(i), where b is a constant. It is given that this palne passes through (2, -3, 5).
So, 5 = b
Substituting this value in (i), we get z = 5, which is the required equation of the plane.
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Question 52 Marks
Write the general equation of a plane parallel to X-axis.
Answer
The general equation of a plane is
ax + by + cz + d = 0 .....(i)
This plane is parallel to the X-axis.
It means that this plane passes through the point (0, y, z). So,
a(0) + by + cz + d = 0
⇒ by + cz + d = 0
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Question 62 Marks
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
Answer
For x-intercepts, put y = 0 and z = 0 in the given eqution.
Then, we get
2x + 0 - 0 = 5
⇒ 2x = 5
$\Rightarrow\text{x}=\frac{5}{2}$
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Question 72 Marks
Write the equation of the plane $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$ in scalar product from.
Answer
The given equation of the plane is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}}$
So, the plane passes through the vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ whose normal vector is $\vec{\text{b}}\times\vec{\text{a}}$
(It means that $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{a}}$)
So, the eqution of the plane in scalar product from is
$(\vec{\text{r}}-\vec{\text{a}}).\vec{\text{n}}=0$
$\Rightarrow(\vec{\text{r}}-\vec{\text{a}}).(\vec{\text{b}}\times\vec{\text{c}})=0$
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Question 82 Marks
Find the vector equation of a plane which is at a distance of 5 unit from the origin and which is normal to the vector $\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Answer
It is given that the normal vector, $\vec{\text{n}}=\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}$
Now, $\text{n}=\frac{\vec{\text{n}}}{|\vec{\text{n}}|}$
$=\frac{\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}}{\sqrt{1+4+4}}$
$=\frac{\hat{\text{i}}-2\hat{\text{j}}-2\hat{\text{k}}}{3}$
$=\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}$
The equation of a plane in normal form is
$\vec{\text{r}}\cdot{\text{n}}={\text{d}}$ (where d is the distance of the plane from the origin)
Substituting $\text{n}=\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}$ and d = 5
Here,
$\vec{\text{r}}\cdot\Big(\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}\Big)=5$
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Question 92 Marks
Find the cartesian form of the equation of a plane whose vector equation is:
$\vec{\text{r}}\cdot\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$ 
Answer
Given the vector equation of a plane,
$\vec{\text{r}}\cdot\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$
Let, $\vec{\text{r}}=\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)$
$\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)\big(12\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}\big)+5=0$
$(\text{x})(12)+(\text{y})(-3)+(\text{z})(4)=0$
$12\text{x}-3\text{y}+4\text{z}+5=0$
Cartesian form of the equation of the plane of the plane is given by
$12\text{x}-3\text{y}+4\text{z}+5=0$
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Question 102 Marks
Find the equation of the plane passing through the following point:
(2, 3, 4), (-3, 5, 1) and (4, -1, 2)
Answer
The equation of the plane passing through points (2, 3, 4), (-3, 5, 1) and (4, -1, 2) is given by,
$\begin{vmatrix}\text{x}-2&\text{y}-3&\text{z}-4\\-3-2&5-3&1-4\\4-2&-1-3&2-4\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-3&\text{z}-4\\-5&2&-3\\2&-4&-2\end{vmatrix}=0$
$\Rightarrow-16(\text{x}-2)-16(\text{y}-3)+16(\text{z}-4)=0$
$\Rightarrow(\text{x}-2)+(\text{y}-3)-(\text{z}-4)=0$
$\Rightarrow\text{x}+\text{y}-\text{z}=1$
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Question 112 Marks
Write the value of k for which the line $\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{\text{k}}$ is perpendicular to the normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})=4.$
Answer
Direction ratios of the given line $\frac{\text{x}-1}{2}=\frac{\text{y}-1}{3}=\frac{\text{z}-1}{\text{k}}$ are proportional to 2, 3, k.
Direction ratios of the normal to the plane $\vec{\text{r}}.(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})=4$ are 2, 3, 4.
Given that these two are perpendicular.
⇒ (2) (2) + (3) (3) + (k) (4) (Because a1a2 + b1b2 + c1c2 = 0)
⇒ 4 + 9 + 4k = 0
⇒ 13 + 4k = 0
$\Rightarrow\text{k}=\frac{-13}{4}$.
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Question 122 Marks
Find the equation of the plane passing through the following point:
(1, 1, 1), (1, -1, 2) and (-2, -2, 2)
Answer
The equation of the plane passing through points (1, 1, 1), (1, -1, 2) and (-2, -2, 2) is given by,
$\begin{vmatrix}\text{x}-1&\text{y}-1&\text{z}-1\\1-1&-1-1&2-1\\-2-1&-2-1&2-1\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-1&\text{y}-1&\text{z}-1\\0&-2&1\\-3&-3&1\end{vmatrix}=0$
$\Rightarrow1(\text{x}-1)-3(\text{y}-1)-16(\text{z}-1)=0$
$\Rightarrow\text{x}-1-3\text{y}+3-6\text{z}+6=0$
$\Rightarrow\text{x}-3\text{y}-6\text{z}+8=0$
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Question 132 Marks
Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.
Answer
We know that the distance of the point (x1, y1, z1) from the plane ax + by + cz + d = 0 is given by
$\frac{|\text{ac}_1+\text{by}_1+\text{cz}_1+\text{d}|}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
So, the required distance
$=\frac{|2(0)-3(0)+6(0)+21|}{\sqrt{2^2+(-3)^2+6^2}}$
$=\frac{|21|}{\sqrt{4+9+36}}$
$=\frac{21}{7}$
$=3\text{ units}$
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Question 142 Marks
Write the distance of the plane $\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=12$ form the origin.
Answer
The given equation of the plane is
$\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=12$ or $\vec{\text{r}}.\vec{\text{n}}=-6$, where $\vec{\text{n}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{4+1+4}=3$
For reducing the given equation to normal form, we need to divide both sides by $|\vec{\text{n}}|$.
Then, we get $\vec{\text{r}}.\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{12}{|\vec{\text{n}}|}$
$=\vec{\text{r}}.\Big(\frac{2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}}{3}\Big)=\frac{12}{3}$
$\Rightarrow\vec{\text{r}}.\Big(\frac{2}{3}\hat{\text{i}}+\frac{1}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}\Big)=4,\ ...(1)$
The eqution of the plane normal from is.
$\vec{\text{r}}.\text{n}=\text{d}\ ....(2)$
(where d is the distance of the plane from the origin)
Comparing (1) and (2),
Length of the perpendicular from the origin to the plane = d = 4 units.
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Question 152 Marks
Write a vector normal to the plane $\vec{\text{r}}=\text{l}\vec{\text{b}}+\text{m}\vec{\text{c}}$ .
Answer
The equation of the given plane is
$\vec{\text{r}}=\text{l}\vec{\text{b}}+\text{m}\vec{\text{c}}$
So, the plane passes parallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the vector normal to the plane is $\vec{\text{b}}\times\vec{\text{c}}$.
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Question 162 Marks
Write the position vector of the point where the line $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$  meets the plane $\vec{\text{r}}.\vec{\text{n}}=0$.
Answer
The given equation of the plane is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\ ....(1)$
Given equation of the plane is
$\vec{\text{r}}.\vec{\text{n}}=0$
$\Rightarrow\big(\vec{\text{a}}+\lambda\vec{\text{b}}\big)\vec{\text{n}}=0$ [ From (1) ]
$\Rightarrow\vec{\text{a}}.\vec{\text{n}}+\lambda\vec{\text{b}}.\vec{\text{n}}=0$
$\Rightarrow\lambda=-\Big(\frac{\vec{\text{a}}.\vec{\text{n}}}{\vec{\text{b}}.\vec{\text{n}}}\Big)$
Substituting this in (1), we get
$\vec{\text{r}}=\vec{\text{a}}-\Big(\frac{\vec{\text{a}}.\vec{\text{n}}}{\vec{\text{b}}.\vec{\text{n}}}\Big)\vec{\text{b}},$ which is the required position vector that lies both on the line and the plane.
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Question 172 Marks
Find the equation of the plane passing through the following point:
(0, -1, 0), (3, 3, 0) and (1, 1, 1)
Answer
The equation of the plane passing through points (0, -1, 0), (3, 3, 0) and (1, 1, 1) is given by,
$\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\3-0&3+1&0-0\\1-0&1+1&1-0\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-0&\text{y}+1&\text{z}-0\\3&4&0\\1&2&1\end{vmatrix}=0$
$\Rightarrow4\text{x}-3(\text{y}+1)+2\text{z}=0$
$\Rightarrow4\text{x}-3\text{y}+2\text{z}=3$
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Question 182 Marks
Answer each of the following questions in one word or one sentence or as per exact requirement of the quetion:
Find the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$
Answer
The required plane passes through $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ and is parallel to the plane $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2.$
So, it is normal to the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ which is normal to the given plane.
Hence, the vector equation of the required plane is
$\big[\vec{\text{r}}-\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0\big]\ \big[\big(\vec{\text{r}}-\vec{\text{a}}\big).\vec{\text{n}}=0\big]$
$\Rightarrow\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}$
Thus, the vector equation of the required plane is $\vec{\text{r}}.(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=\text{a}+\text{b}+\text{c}.$
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Question 192 Marks
Find the cartesian form of the equation of a plane whose vector equation is:
$\vec{\text{r}}\cdot\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$ 
Answer
Here, equation of the plane is,
$\vec{\text{r}}\cdot\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$
Let $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$ then
$\big(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}\big)\big(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)=9$
$(\text{x})(-1)+(\text{y})(1)+(\text{z})(2)=9$
$-\text{x}+\text{y}+2\text{z}=9$
Cartesian form of the equation of the plane is,
$-\text{x}+\text{y}+2\text{z}=9$
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Question 202 Marks
Find the vector equation of a plane passing throught a point with position $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and perpendicular to the vector $4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ 
Answer
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{n}}=4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},$ we get
$\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=8-2-3$
$\Rightarrow\vec{\text{r}}\cdot\big(4\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\big)=3$
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Question 212 Marks
Write the value of k for which the planes x - 2y + kz = 4 and 2x + 5y - z = 9 are perpendicular.
Answer
We know that the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are prependicular to each of a1a2 + b1b2 + c1c2 = 0
The given planes are x - 2y + kz = 4 and 2x + 5y - z = 9
⇒ a1 = 1; b1 = -2; c1 = 5; a2 = 2; b2 = 5; c2 = -1
It is given that the given planes are perpendicular.
⇒ a1a2 + b1b2 + c1c2 = 0
⇒ (1)(2) + (-2)(5) + (k)(-1) = 0
⇒ 2 - 10 - k = 0
⇒ -8 - k = 0
⇒ k = -8
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Question 222 Marks
Write the equation of the plane corntaining the lines $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$  and $\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{c}}$.
Answer
The given equation of the plane is
 $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$  and $\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ and parallel to the vector $\vec{\text{b}}$ and $\vec{\text{c}}$.
So, the plane passes through the vector $\vec{\text{a}}$ whose normal vector is $\vec{\text{b}}\times\vec{\text{a}}$
(It means that $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{a}}$)
So, the eqution of the plane in scalar product from is
$(\vec{\text{r}}-\vec{\text{a}}).\vec{\text{n}}=0$
$\Rightarrow(\vec{\text{r}}-\vec{\text{a}}).(\vec{\text{b}}\times\vec{\text{c}})=0$
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Question 232 Marks
Find the vector equation one of following plane.
x + y = 3
Answer
Given, equation of plane is,
x + y = 3
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}})=3$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}})=3$
So,
Vector equation of the plane is $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}})=3$
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Question 242 Marks
Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
Answer
The equation of the plane passing through (a, 0, 0), (0, b, 0) and (0, 0, c) is
$\begin{vmatrix}\text{x}-\text{a} & \text{y}-0&\text{z}-0 \\ 0-\text{a} & \text{b}-0 & 0 - 0 \\ 0-\text{a}&0-0&\text{C}-0 \end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}\text{x}-\text{a} & \text{y}&\text{z} \\ -\text{a} & \text{b} & 0 - 0 \\ -\text{a}& 0 &\text{C} \end{vmatrix}=0$
$\Rightarrow\ \text{bc}(\text{x}-\text{a})+\text{acy}+\text{abz}=0$
$\Rightarrow\ \text{bcx}+\text{acy}+\text{abz}=\text{abc}$
Dividing the equationg by abc, we get
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1$
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Question 252 Marks
Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y − z = 7.
Answer
Let the equation of a plane parallel to the given plane be
3x + 2y - z = k ....(1)
This passes through (2, -1, 1).
So, 3(2) + 2(-1) - (1) = k
k = 3
Substituting this in(1),
We get,
3x + 2y - z = 3, which is the equation of the required plane.
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Question 262 Marks
Write the angle between the line $\frac{\text{x}-1}{2}=\frac{\text{y}-2}{1}=\frac{\text{z}+3}{-\text{2}}$ and the plane x + y + 4 = 0.
Answer
The given line is parallel to the vector $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and the given plane is normal to the vector $\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$.
We know that the angle $\theta$ between the line and the plane is given by
$\sin\theta=\frac{\vec{\text{b}}.\vec{\text{n}}}{|\vec{\text{b}}||\vec{\text{n}}|}$
$=\frac{\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)\big(\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}\big)}{|\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}||\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}|}$
$=\frac{1+2+0}{\sqrt{1+4+4}\sqrt{1+1+0}}$
$=\frac{3}{3\sqrt{2}}$
$=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\sin^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=45^\circ$
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Question 272 Marks
Write the equation of the plane parallel to the YOZ- plane and passing through (-4, 1, 0).
Answer
The equation of the plane parallel ot the plane YOZ is x = b .....(i), where b is a constant. It is given that plane passes throught (-4, 1, 0). So, -4 = b
Substituting this value in (i), we get x = -4, which is the required equation of the plane.
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Question 282 Marks
Find the vector equation of the plane which is at a distance of 5 units from the orgin and its normal vector is $2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}.$
Answer
Given:

Normal vector, $\hat{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$

Perpendicular distance, d = 5 units

The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector $\hat{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$ is as follows:

$\vec{\text{r}}.\hat{\text{n}}=\text{d}$

$\vec{\text{r}}.\big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\big)=5.$

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