Fill In The Blanks[1 Marks ]

Question 11 Mark

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If $\vec{\text{a}}$ is any non-zero vector, then $(\vec{\text{a}}\cdot\vec{\text{i}})\vec{\text{i}}+(\vec{\text{a}}\cdot\vec{\text{j}})\vec{\text{j}}+(\vec{\text{a}}\cdot\vec{\text{k}})\vec{\text{k}}$ equal ________.

Answer

If $\vec{\text{a}}$ is any non-zero vector, then $(\vec{\text{a}}\cdot\vec{\text{i}})\vec{\text{i}}+(\vec{\text{a}}\cdot\vec{\text{j}})\vec{\text{j}}+(\vec{\text{a}}\cdot\vec{\text{k}})\vec{\text{k}}$ equal $\vec{\text{a}}={\text{a}}_1\hat{{\text{i}}}+{\text{a}}_2\hat{{\text{j}}}+{\text{a}}_3\hat{{\text{k}}}.$Solution:
$\vec{\text{a}}={\text{a}}_1\hat{{\text{i}}}+{\text{a}}_2\hat{{\text{j}}}+{\text{a}}_3\hat{{\text{k}}}$
Taking dot producrt with $\hat{\text{i}},$ we get ${\text{a}}_1\cdot\hat{{\text{i}}}={\text{a}}_1\hat{{\text{i}}}+0+0$
$\Rightarrow{\text{a}}_1=\vec{\text{a}}\cdot\hat{{\text{i}}}$
Similarly taking dot product with $\hat{{\text{j}}}$ and $\hat{{\text{k}}},$ we get
${\text{a}}_1=\vec{\text{a}}\cdot\hat{{\text{j}}}=\text{a}_2$ and $\vec{\text{a}}\cdot\hat{{\text{k}}}={\text{a}}_3$
$\therefore\vec{\text{a}}={\text{a}}_1\hat{{\text{i}}}+{\text{a}}_2\hat{{\text{j}}}+{\text{a}}_3\hat{{\text{k}}}$ $=(\vec{\text{a}}\cdot\vec{\text{i}})\vec{\text{i}}+(\vec{\text{a}}\cdot\vec{\text{j}})\vec{\text{j}}+(\vec{\text{a}}\cdot\vec{\text{k}})\vec{\text{k}}$

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Question 21 Mark

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If $\vec{\text{r}}\cdot\vec{\text{a}}=0,\vec{\text{r}}\cdot\vec{\text{b}}=0,$ and $\vec{\text{r}}\cdot\vec{\text{c}}=0$ for some non-zero vector $\vec{\text{r}},$ then the value of $\vec{\text{a}}(\vec{\text{b}}\times\vec{\text{c}})$ is _______.

Answer

If $\vec{\text{r}}\cdot\vec{\text{a}}=0,\vec{\text{r}}\cdot\vec{\text{b}}=0,$ and $\vec{\text{r}}\cdot\vec{\text{c}}=0$ for some non-zero vector $\vec{\text{r}},$ then the value of $\vec{\text{a}}(\vec{\text{b}}\times\vec{\text{c}})$ is 0.Solution:
Since, $\vec{\text{r}}$ is a non-zero vector, so we can say that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are in a same plane.
$\vec{\text{a}}(\vec{\text{b}}\times\vec{\text{c}})=0$
[Since, angle between $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are zero i.e., $\theta=0]$

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Question 31 Mark

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The value of the expression $|\vec{\text{a}}\times\vec{\text{b}}|^2+(\vec{\text{a}}\cdot\vec{\text{b}})^2$ is ________.

Answer

The value of the expression $|\vec{\text{a}}\times\vec{\text{b}}|^2+(\vec{\text{a}}\cdot\vec{\text{b}})^2$ is $|\vec{\text{a}}|^2|\vec{\text{b}}|^2.$Solution:
$|\vec{\text{a}}\times\vec{\text{b}}|^2+(\vec{\text{a}}\cdot\vec{\text{b}})^2=|\vec{\text{a}}|^2|\vec{\text{b}}|^2\sin^2\theta+(\vec{\text{a}}\cdot\vec{\text{b}})^2$ $=|\vec{\text{a}}|^2|\vec{\text{b}}|^2(1-\cos^2\theta)+(\vec{\text{a}}\cdot\vec{\text{b}})^2$ $=|\vec{\text{a}}|^2|\vec{\text{b}}|^2-|\vec{\text{a}}|^2|\vec{\text{b}}|^2\cos^2\theta+(\vec{\text{a}}\cdot\vec{\text{b}})^2$ $=|\vec{\text{a}}|^2|\vec{\text{b}}|^2-(\vec{\text{a}}\cdot\vec{\text{b}})^2+(\vec{\text{a}}\cdot\vec{\text{b}})^2$ $|\vec{\text{a}}|^2|\vec{\text{b}}|^2+(\vec{\text{a}}\cdot\vec{\text{b}})^2=|\vec{\text{a}}|^2|\vec{\text{b}}|^2$

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Question 41 Mark

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The values of k for which $|\text{k}\vec{\text{a}}|<|\vec{\text{a}}|$ and $\text{k}\vec{\text{a}}=\frac{1}{2}\vec{{\text{a}}}$ is a parallel to $\vec{\text{a}}$ holds true are _________.

Answer

The values of k for which $|\text{k}\vec{\text{a}}|<|\vec{\text{a}}|$ and $\text{k}\vec{\text{a}}=\frac{1}{2}\vec{{\text{a}}}$ is a parallel to $\vec{\text{a}}$ holds true are $\text{k}\vec{\text{a}}+\frac{1}{2}\vec{\text{a}}.$
Solution:
We have, $|\text{k}\vec{\text{a}}|<|\vec{\text{a}}|$ and $\text{k}\vec{\text{a}}=\frac{1}{2}\vec{{\text{a}}}$ parallel to $\vec{\text{a}}$
$\therefore|\text{k}\vec{\text{a}}|<|\vec{\text{a}}|\Rightarrow\text{k}|\vec{\text{a}}|<|\vec{\text{a}}|$
$\Rightarrow|\text{k}|<1\Rightarrow-1<\text{k}<1$
Also, since $\text{k}\vec{\text{a}}=\frac{1}{2}\vec{{\text{a}}}$ is a parallel to $\vec{\text{a}},$ then we see that at $\text{k}\vec{\text{a}}=\frac{1}{2}\text{k}\vec{{\text{a}}}+\frac{1}{2}\vec{\text{a}}$ becomes a null vector and then it will then it will not be parallel to $\vec{\text{a}}.$
So, $\text{k}\vec{\text{a}}=\frac{1}{2}\vec{{\text{a}}}$ is parallel to $\vec{{\text{a}}}$ holds true when $\text{k}\epsilon[-1,1],\text{k}\neq\frac{-1}{2}.$

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Question 51 Mark

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The vector $\vec{\text{a}}+\vec{\text{b}}$ bisects the angle between the non-collinear vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if _________.

Answer

The vector $\vec{\text{a}}+\vec{\text{b}}$ bisects the angle between the non-collinear vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if $\vec{\text{a}}=\vec{\text{b}}.$Solution:
If vector $\vec{\text{a}}=\vec{\text{b}}$ bisects the angle between the non-collinear vectors, then $\vec{\text{a}}\cdot(\vec{\text{a}}+\vec{\text{b}})=|\vec{\text{a}}|\cdot|\vec{\text{a}}+\vec{\text{b}}|\cos\theta$ $\vec{\text{b}}\cdot(\vec{\text{a}}+\vec{\text{b}})=\text{b}\sqrt{\text{a}^2+\text{b}^2}\cos\theta$ $\Rightarrow\cos\theta=\frac{\vec{\text{a}}\cdot(\vec{\text{a}}+\vec{b})}{\text{a}\sqrt{\text{a}^2+\text{b}^2}}\ ....(\text{i})$ And $\vec{\text{b}}\cdot({\vec{\text{a}}}+\vec{\text{b}})=|\vec{\text{b}}|\cdot|\vec{\text{a}}+\vec{\text{b}}|\cos\theta$ [since, $\theta$ should be same] $\Rightarrow\cos\theta=\frac{\vec{\text{b}}\cdot(\vec{\text{c}}+\vec{b})}{\text{b}\sqrt{\text{a}^2+\text{b}^2}}\ ....(\text{ii})$ From equations (i) and (ii), we get: $\Rightarrow\frac{\vec{\text{a}}\cdot(\vec{\text{a}}+\vec{b})}{\text{a}\sqrt{\text{a}^2+\text{b}^2}}=\frac{\vec{\text{b}}\cdot(\vec{\text{c}}+\vec{b})}{\text{b}\sqrt{\text{a}^2+\text{b}^2}}$ $\Rightarrow\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{\vec{\text{b}}}{|\vec{\text{b}}|}$ $\therefore\vec{\text{a}}=\vec{\text{b}}$ Thus, the vector $\vec{\text{a}}+\vec{\text{b}}$ bisects the angle between the non-collinear vector $\vec{\text{a}}$ and $\vec{\text{b}},$ if $\vec{\text{a}}$ and $\vec{\text{b}}$ are equal vectors.

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Question 61 Mark

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If $|\vec{\text{a}}\times\vec{\text{b}}|^2+|\vec{\text{a}}\cdot\vec{\text{b}}|^2=144$ and $|\vec{\text{a}}|=4,$ then $|\vec{\text{b}}|^2$ is equal to ________.

Answer

If $|\vec{\text{a}}\times\vec{\text{b}}|^2+|\vec{\text{a}}\cdot\vec{\text{b}}|^2=144$ and $|\vec{\text{a}}|=4,$ then $|\vec{\text{b}}|^2$ is equal to 3.Solution:
$|\vec{\text{a}}\times\vec{\text{b}}|^2+|\vec{\text{a}}\cdot\vec{\text{b}}|^2=144$ $|\vec{\text{a}}|^2|\vec{\text{b}}|^2\sin^2\theta+|\vec{\text{a}}|^2|\vec{\text{b}}|^2\cos^2\theta=144$ $\Rightarrow|\vec{\text{a}}|^2|\vec{\text{b}}|^2=144$ $\Rightarrow|\vec{\text{a}}|^2|\vec{\text{b}}|^2=12$ $\Rightarrow4|\vec{\text{b}}|^2=12$ $\Rightarrow|\vec{\text{b}}|^2=3$

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Question 71 Mark

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The vectors $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}-2\hat{\text{k}}$ are the adjacent sides of a parallelogram. The angle between its diagonals is _________.

Answer

The vectors $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}-2\hat{\text{k}}$ are the adjacent sides of a parallelogram. The angle between its diagonals is $\theta=\frac{\pi}{4}.$Solution:
We have, $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}-2\hat{\text{k}}$ $\therefore\vec{\text{a}}+\vec{\text{b}}=2\hat{\text{i}}-2\hat{\text{j}}$ and $\vec{\text{a}}-\vec{\text{b}}=4\hat{\text{i}}-2\hat{\text{j}}+4\vec{\text{k}}$ Now, let $\theta$ is the acute angle between the diagonals $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}.$ $\therefore\cos\theta=\frac{(\vec{\text{a}}+\vec{\text{b}})\cdot(\vec{\text{a}}-\vec{\text{b}})}{|(\vec{\text{a}}+\vec{\text{b}})||(\vec{\text{a}}-\vec{\text{b}})|}$ $=\frac{(2\vec{\text{i}}-2\vec{\text{j}})\cdot(4\vec{\text{i}}-2\vec{\text{j}}+4\vec{\text{k}})}{\sqrt{8}\sqrt{16+4+16}}$ $=\frac{8+4}{2\sqrt{2}\cdot6}=\frac{1}{\sqrt{2}}$ $\therefore\theta=\frac{\pi}{4}$ $\Big[\because\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}\Big]$

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