Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blanks.
The vector equation of the line through the points (3, 4, -7) and (1, -1, 6) is __________.

Answer

The vector equation of the line through the points (3, 4, -7) and (1, -1, 6) is $(\text{x}-3)\hat{\text{i}}(\text{y}-4)\hat{\text{j}}+(\text{z}+7)\hat{\text{k}}=\lambda(-2\hat{\text{i}}-5\hat{\text{j}}+13\hat{\text{k}}).$
Solution:
We know that, vector equation of a line passes through two points $\vec{\text{a}}$ and $\vec{\text{b}}$ is representes by $\vec{\text{r}}=\vec{\text{a}}+\lambda(\vec{\text{b}}-\vec{\text{a}})$
Here, $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}-7\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+6\hat{\text{k}}$
So, the required equation is
$\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=3\hat{\text{i}}+4\hat{\text{j}}-7\hat{\text{k}}+\lambda(-2\hat{\text{i}}-5\hat{\text{j}}+13\hat{\text{k}})$
$\Rightarrow(\text{x}-3)\hat{\text{i}}(\text{y}-4)\hat{\text{j}}+(\text{z}+7)\hat{\text{k}}=\lambda(-2\hat{\text{i}}-5\hat{\text{j}}+13\hat{\text{k}})$

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Question 21 Mark

Fill in the blanks.
The cartesian equation of the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=2$ is ________.

Answer

The cartesian equation of the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=2$ is $\text{x}+\text{y}-\text{z}=2.$
Solution:
We have, $\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=2$
$\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})=2$
$\Rightarrow\text{x}+\text{y}-\text{z}=2$
This is the required cartesian from.

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Question 31 Mark

Fill in the blanks.
A plane passes through the points (2, 0, 0) (0, 3, 0) and (0, 0, 4). The equation of plane is __________.

Answer

A plane passes through the points (2, 0, 0) (0, 3, 0) and (0, 0, 4). The equation of plane is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1.$Solution:
We know that, equation of the plane that cut the coordinate axes at (a, 0, 0), (0, b, 0) and (0, 0, c) is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1.$

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Question 41 Mark

Answer

The position vectors of two points A and B are $\overrightarrow{\text{OA}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$ and $\overrightarrow{\text{OB}}=2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}},$ respectively. The position vector of a point P which divides the line segment joining A and B in the ratio 2 : 1 is $2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}.$
Solution:
The position vector are
$\overrightarrow{\text{OA}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$\overrightarrow{\text{OB}}=2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}},$
Let P divide AB, in the ratio such that 2 : 1
$\text{P}=\frac{\text{m}\vec{\text{b}}+\text{n}\vec{\text{a}}}{\text{a+b}}$
$\text{P}=\frac{2(2\vec{\text{i}}-\vec{\text{j}}+2\vec{\text{k}})+1(2\vec{\text{i}}-\vec{\text{j}}-\vec{\text{k}})}{2+1}$
$\text{P}=\frac{6\vec{\text{i}}-3\vec{\text{j}}+3\vec{\text{k}}}{3}$
$=\text{P}=2\vec{\text{i}}-\vec{\text{j}}+\vec{\text{k}}$

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Question 51 Mark

Fill in the blanks.
The vector equation of the line $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}$ is _________.

Answer

The vector equation of the line $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}$ is $(\text{x}-5)\hat{\text{i}}+(\text{y}+5)\hat{\text{j}}+(\text{z}-6)\hat{\text{z}}=\lambda(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}).$Solution:
We have, $\vec{\text{a}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}$ So, the vector equation will be $\vec{\text{r}}=(5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}})+\lambda(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}})$ $\Rightarrow(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})-(5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}})=\lambda(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}})$ $(\text{x}-5)\hat{\text{i}}+(\text{y}+5)\hat{\text{j}}+(\text{z}-6)\hat{\text{z}}=\lambda(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}})$

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Question 61 Mark

Fill in the blanks.
The direction cosines of the vector $2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ are ________.

Answer

The direction cosines of the vector $2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ are $\frac{2}{3},\frac{2}{3},\frac{-1}{3}.$Solution:
The direction cosines of the vector $({\text{a}}\hat{\text{i}}+{\text{b}}\hat{\text{j}}+{\text{c}}\hat{\text{k}})$ are given by $\frac{{\text{a}}}{\sqrt{{\text{a}}^2+{\text{b}}^2+{\text{c}}^2}},\frac{{\text{b}}}{\sqrt{{\text{a}}^2+{\text{b}}^2+{\text{c}}^2}},\frac{{\text{c}}}{\sqrt{{\text{a}}^2+{\text{b}}^2+{\text{c}}^2}}$ Therefore, the direction cosines of the vector $2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ are $\frac{2}{\sqrt{2^2+2^2+(-1)^2}},\frac{2}{\sqrt{2^2+2^2+(-1)^2}},\frac{-1}{\sqrt{2^2+2^2+(-1)^2}}$ Or $\frac{2}{\sqrt{4+4+1}},\frac{2}{\sqrt{4+4+1}},\frac{-1}{\sqrt{4+4+1}}$ i.e., Or $\frac{2}{3},\frac{2}{3},\frac{-1}{3}.$

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Maths Fill In The Blanks[1 Marks ]

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