2 Marks Questions

Question 12 Marks

If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-coplanar vectors, then find the value of $\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$

Answer

We have
$\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$
$\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}+\frac{\big[\vec{\text{b}}\vec{ \text{a}}\vec{\text{c}}\big]}{\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]}$ (By definition of scalar triple product)
$=\frac{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}+\frac{-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}{\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]}$ (Change of cyclic order of vectors changes the sign of the scalar triple product)
$=1-1$
$=0$

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Question 22 Marks

Find the volume of the parallelopiped with its edges represented by the vectors
$\hat{\text{i}}+\hat{\text{j}},\hat{\text{i}}+2\hat{\text{j}}$ and $\hat{\text{i}}+\hat{\text{j}}+\pi{\text{k}}.$

Answer

Let:
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}$
$\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\pi\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjecent edges are $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ is equal to $\big|\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]\big|.$
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\begin{vmatrix}1&1&0\\1&2&0\\1&1&\pi \end{vmatrix}$
$=1(2\pi-0)-1(\pi-0)+0(1-2)$
$=2\pi-\pi=\pi$
$\therefore$ volume $=\big|\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]\big|=|\pi|=\pi$ cubic units

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Question 32 Marks

Write the value of $\big[2\hat{\text{i }} 3\hat{\text{j }}4\hat{\text{k}}\big].$

Answer

We have
$\big[2\hat{\text{i }}3\hat{\text{j }}4\hat{\text{k}}\big]$
$=\big(2\hat{\text{i}}\times3\hat{\text{j}}\big).4\hat{\text{k}}$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$=6\hat{\text{k}}.4\hat{\text{k}}$
$=24$

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Question 42 Marks

Write the value of $\big[\hat{\text{i}}-\hat{\text{j }}\hat{\text{j}}-\hat{\text{k }}\hat{\text{k}}-\hat{\text{i}}\big].$

Answer

We have
$\big[\hat{\text{i}}-\hat{\text{j }}\hat{\text{j}}-\hat{\text{k }}\hat{\text{k}}-\hat{\text{i}}\big]=\big[\big(\hat{\text{i}}-\hat{\text{j}}\big)\times\big(\hat{\text{j}}-\hat{\text{k}}\big)\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$\big[\big(\hat{\text{i}}\times\hat{\text{j}}\big)-\big(\hat{\text{i}}\times\hat{\text{k}}\big)-\big(\hat{\text{j}}\times{\hat{\text{j}}}\big)+\big(\hat{\text{j}}\times\hat{\text{k}}\big)\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$=\big[\hat{\text{k}}+\hat{\text{j}}+\hat{\text{i}}\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$=\big[\big(\hat{\text{k}}.\hat{\text{k}}\big)-\big(\hat{\text{k}}.\hat{\text{i}}\big)+\big(\hat{\text{j}}.\hat{\text{k}}\big)-\big(\hat{\text{j}}.\hat{\text{i}}\big)+\big(\hat{\text{i}}.\hat{\text{k}}\big)-\big(\hat{\text{i}}.\hat{\text{i}}\big)\big]$
$=1-0+0-0+0-1=0$

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Question 52 Marks

If $\vec{\text{a}},\vec{\text{b}}$ are non-collinear vectors, then find the value of $\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{i}}\big]\hat{\text{i}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{j}}\big]\hat{\text{j}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{k}}\big]\hat{\text{k}}.$

Answer

For any vector $\vec{\text{r}},$ we have
$\big(\vec{\text{r}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{r}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{r}}.\hat{\text{k}}\big)\hat{\text{k}}=\vec{\text{r}}$
Replacing $\vec{\text{r}}$ by $\vec{\text{a}}\times\vec{\text{b}},$ we have
$\big[(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{i}}\big]+\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{j}}+\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{k}}\big]\hat{\text{k}}=\vec{\text{a}}\times\vec{\text{b}}$
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{i}}\big]\hat{\text{i}}\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{j}}\big]\hat{\text{j}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{k}}\big]\hat{\text{k}}=\vec{\text{a}}\times\vec{\text{b}}$

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Question 62 Marks

Show that the following triads of vectors are coplanar:
$\vec{\text{a}}=-4\hat{\text{i}}-6\hat{\text{j}}-2\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}},\vec{\text{c}}=-8\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$

Answer

We know that three vectors $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$ are coplanar iff their scalar triple product is zero i.e. $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}-4&-6&-2\\-1&4&3\\-8&-1&3 \end{vmatrix}\\$
$=-4(12+3)+6(-3+24)-2(1+32)$
$= -60 + 126 -66$
$= 0$
Hence, the given vectors are coplanar.

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Question 72 Marks

Find the values of 'a' for which the vectors
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},\vec\beta={\text{a}}\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ and $\vec\gamma=\hat{\text{i}}+2\hat{\text{j}}+\text{a}\hat{\text{k}}$ are coplanar.

Answer

Given:
$\vec\alpha=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\vec\beta={\text{a}}\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\vec\gamma=\hat{\text{i}}+2\hat{\text{j}}+\text{a}\hat{\text{k}}$
We know that three vectors $\vec\alpha,\vec\beta,\vec\gamma$ are coplanar if their scalar product is zero.
$\therefore\big[\vec\alpha \vec \beta \vec\gamma\big]=0$
$\begin{vmatrix}1&2&1\\\text{a}&1&2\\1&2&\text{a} \end{vmatrix}=0$
$\Rightarrow1(\text{a}-4)-2(\text{a}^2-2)+1(2\text{a}-1)=0$
$\Rightarrow-2\text{a}^2+3\text{a}-1=0$
$\Rightarrow2\text{a}^2-3\text{a}+1=0$
$\Rightarrow(\text{a}-1)(2\text{a}-1)=0$
$\Rightarrow\text{a}=1,\frac{1}{2}$

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Question 82 Marks

Find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big), $ if $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.

Answer

The given vectors are $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\vec{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Now,
$\vec{\text{b}}\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-1&2&1\\3&1&2 \end{vmatrix}=3\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)=\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big).\big(3\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}\big)$
$=2\times3+1\times5=3\times(-7)$
$=6+5-21=-10$

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Question 92 Marks

Show that the following triads of vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+7\hat{\text{k}},\vec{\text{c}}=5\hat{\text{i}}+6\hat{\text{j}}+5\hat{\text{k}}$

Answer

We know that three vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff their scalar triple product is zero i.e. $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}1&2&-1\\3&2&7\\5&6&5 \end{vmatrix}$
$=1(1-42)-2(15-35)-1(18-10)$
$=-60+126-66$
$=0$
Hence, the given vectors are coplanar.

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Question 102 Marks

Evaluate the following:$\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]$

Answer

We have,
$\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat {\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]\\=(2\hat{\text{i}}\times\hat{\text{j}}).\hat{\text{k}}+(\hat{\text{i}}\times\hat{\text{k}}).2\hat{\text{i}}$
$=2\hat{\text{k}}.\hat{\text{k}}+(-\hat{\text{j}}).\hat{\text{j}}+(-\hat{\text{i}}).2\hat{\text{i}}$
$=2-1-2$
$=-1$
Therefore, $\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]=-1$

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Question 112 Marks

Evaluate the following:
$\big[\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{j}}\hat{\text{k}}\hat{\text{i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]$

Answer

We have,
$\big[\hat{\text{i}}\hat{\text{ j}}\hat{\text{ k}}\big]+\big[\hat{\text{j}}\hat{\text{ k}}\hat{\text{ i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]$
$=(\hat{\text{i}}\times\hat{\text{j}}).\hat{\text{k}}(\hat{\text{j}}\times\hat{\text{k}}).\hat{\text{i}}+(\hat{\text{k}}\times\hat{\text{i}}).\hat{\text{j}}$
$=\hat{\text{k}}.\hat{\text{k}}+\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}$
$=1+1+1$
$=3$
Therefore, $\big[\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{j}}\hat{\text{k}}\hat{\text{i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]=3$

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Question 122 Marks

Write the value of $\big[\hat{\text{i}}+\hat{\text{j}}\hat{\text{j}}+\hat{\text{k}}\hat{\text{k}}+\hat{\text{i}}\big].$

Answer

We have
$\big[\hat{\text{i}}+\hat{\text{j}}\hat{\text{j}}+\hat{\text{k}}\hat{\text{k}}+\hat{\text{i}}\big]=\big[\big(\hat{\text{i}}+\hat{\text{j}}\big)\times\big(\hat{\text{j}}+\hat{\text{k}}\big)\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$=\big[\hat{\text{i}}\times\hat{\text{j}}+\hat{\text{i}}\times\hat{\text{k}}+\hat{\text{j}}\times\hat{\text{j}}+\hat{\text{j}}\times\hat{\text{k}}\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$
$=\big[\hat{\text{k}}-\hat{\text{j}}+\hat{\text{i}}\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$
$=\big[\big(\hat{\text{k}}.\hat{\text{k}}\big)+\big(\hat{\text{k}}.\hat{\text{i}}\big)-\big(\hat{\text{j}}.\hat{\text{k}}\big)-\big(\hat{\text{j}}.\hat{\text{i}}\big)+\big(\hat{\text{i}}.\hat{\text{k}}\big)+\big(\hat{\text{i}}.\hat{\text{i}}\big)\big]$
$=1+0-0-0+0+1=2$

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