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MATHS M.C.Q (1 Marks)

Scalar Triple Product · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 15 questions.

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M.C.Q (1 Marks)

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15 questions · timed · auto-graded

Question 11 Mark
If $\vec{\text{a}} $ lies in the plane of vectors $\vec{\text{b}}$ and $\vec{\text{c}},$ then which of the following is correct?
  1. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
  2. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=1$
  3. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=3$
  4. $\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=1$
Answer
  1. ​​​​​​$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
Solution:
If $\vec{\text{a}}$ lies in the plane of vectors $\vec{\text{b}}$ and $\vec{\text{c}},$ then $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ will lie in the same plane, i.e. they will be coplanar.
$\therefore \big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
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Question 21 Mark
If $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}$ and $\vec{\text{c}}=5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}},$ then the volume of the parallelopiped with contermious edges $\vec{\text{a}}+\vec{\text{b}},\vec{\text{b}}+\vec{\text{c}},\vec{\text{c}}+\vec{\text{a}}$ is:
  1. 2
  2. 1
  3. -1
  4. 0
Answer
We have
$\vec{\text{a}}+\vec{\text{b}}=\big(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\big)+\big(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)=5\hat{\text{i}}-7\hat{\text{j}}+10\hat{\text{k}}$
$\vec{\text{b}}+\vec{\text{c}}=\big(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}}\big)\big(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)=8\hat{\text{i}}-7\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{c}}+\vec{\text{a}}=\big(5\hat{\text{i}}-3\hat{\text{j}}-2\hat{\text{k}}\big)+\big(2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}\big)=7\hat{\text{i}}-6\hat{\text{j}}+3\hat{\text{k}}$
We know that the volume of parallelopiped whose three adjacent adges are $\vec{\text{a}}+\vec{\text{b}},\vec{\text{b}}+\vec{\text{c}}$ and $\vec{\text{c}}+\vec{\text{a}}$ is equal to:
We have
$\big[\vec{\text{a}}+\vec{\text{b}}\vec{\text{b}}+\vec{\text{c}}\vec{\text{c}}+\vec{\text{a}}\big]=\begin{vmatrix}5&-7&10\\8&-7&3\\7&-6&3\end{vmatrix}$
$=5(-21+18)+7(24-21)+10(-48+49)$
$=(5\times-3)+(7\times3)+(10\times1)$
$=16$
$\therefore$ volume of parallelopiped $=\big|16\big|=16$
Disclaimer: None of the given option is correct.
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Question 31 Mark
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three non-coplanar vectors, then $\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big[\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{a}}+\vec{\text{c}}\big)\big]$ equals:
  1. $0$
  2. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  3. $2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  4. $-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
Answer
  1. $-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
Solution:
$\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big[\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{a}}+\vec{\text{c}}\big)\big]$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{a}}+\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=0+0+\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]+\big[\vec{\text{b}}\vec{\text{a}}\vec{\text{c}}\big]+0+0+0+\big[\vec{\text{c}}\vec{\text{b}}\vec{\text{a}}\big]+0$
$=-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
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Question 41 Mark
$\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big\{\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{a}}-\vec{\text{b}}-\vec{\text{c}}\big)\big\}$ is equal to:
  1. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big)$
  2. $2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big)$
  3. $3\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big)$
  4. $0$
Answer
  1. $3\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big)$
Solution:
$\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big\{\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{a}}-\vec{\text{b}}-\vec{\text{c}}\big)\big\}$
$=\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big(\vec{\text{a}}\times\vec{\text{a}}-\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{a}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big(-\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{a}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+2\vec{\text{b}}-\vec{\text{c}}\big).\big(-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big[\text{a}\text{b}\text{c}\big]+2\big[\text{a}\text{b}\text{c}\big]$
$=3\big[\text{a}\text{b}\text{c}\big]$
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Question 51 Mark
If $\big[2\vec{\text{a}}+4\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big],$ then $\lambda+\mu=$
  1. 6
  2. -6
  3. 10
  4. 8
Answer
  1. 6
Solution:
We have
$\big[2\vec{\text{a}}+4\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big[\big(2\vec{\text{a}}+4\vec{\text{b}}\big]\times\vec{\text{c}}\big].\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$ (By definition of scalar triple product)
$\Rightarrow\big[\big(2\vec{\text{a}}\times\vec{\text{c}}\big)+\big(4\vec{\text{b}}\times\vec{\text{c}}\big)\big].\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big(2\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{d}}+\big(4\vec{\text{b}}\times\vec{\text{c}}\big).\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big[2\vec{\text{a}}\vec{\text{ c }}\vec{\text{d}}\big]+\big[4\vec{\text{b}}\vec{\text{ c }}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{ c }}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{ c }}\vec{\text{d}}\big]$
$\Rightarrow2\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+4\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$ $\big(\therefore\big[\lambda\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ for any scaler $\lambda\big)$
Comparing both sides, we get
$\lambda=2$
$\mu=4$
$\therefore\lambda+\mu=2+4=6$
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Question 61 Mark
The value of $\big[\vec{\text{a}}-\vec{\text{b}},\vec{\text{b}}-\vec{\text{c}},\vec{\text{c}}-\vec{\text{a}}\big],$ where $\big|\vec{\text{a}}\big|=1,\big|\vec{\text{b}}\big|=5,\big|\vec{\text{c}}\big|=3,$ is:
  1. 0
  2. 1
  3. 6
  4. None of these.
Answer
  1. 0
Solution:
We have
$\big[\vec{\text{a}}-\vec{\text{b}},\vec{\text{b}}-\vec{\text{c}},\vec{\text{c}}-\vec{\text{a}}\big]$
$=\big(\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big).\big(\vec{\text{c}}.\vec{\text{a}}\big)$ (By definition of scalar triple product)
$=\big(\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\vec{\text{b}}-\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{b}}\times{\vec{\text{b}}}-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}-0-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)-\big(\vec{\text{a}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}-\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{a}}-\big(\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{c}}+\big(\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{a}}+\big(\vec{\text{b}}\times\vec{\text{c}}\big).\vec{\text{c}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)\\.\vec{\text{a}}\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\big]-\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{c}}\big]+\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{a}}\big]+\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{c}}\big]-\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$
$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-0+0+0-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]\big)$
$=0$
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Question 71 Mark
Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}},\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ and $\vec{\text{c}}=\text{c}_1\hat{\text{i}}+\text{c}_2\hat{\text{j}}+\text{c}_3\hat{\text{k}}$ be three zero vectors such that $\vec{\text{c}}$ is a unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$ If the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6},$ then $\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}^2$ is equal to:
  1.  $0$
  2. $1$
  3. $\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
  4. $\frac{3}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
Answer
  1. $\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
Solution:
We have
$\begin{vmatrix}\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{c}_1&\text{c}_2&\text{c}_3 \end{vmatrix}^2$
$=\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big]^2$ (By definition of scalar triple product)
$=\big[\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|\big|\vec{\text{c}}\big|\cos0^\circ\big]^2$ $\big(\therefore\vec{\text{a}}\times\vec{\text{b}}$ is parallel to vector $\vec{\text{c}}$ as $\vec{\text{c}}$ is perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}\big)$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\frac{\pi}{6}\big)^2$ $\big(\therefore\big|\vec{\text{c}}\big|=1$ and angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6}\big)$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\frac{1}{2}\big)^2$
$=\frac{1}{4}\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
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Question 81 Mark
For non-zero vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ the relation $\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ holds good, if:
  1. $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=0$
  2. $\vec{\text{a}}.\vec{\text{b}}=0=\vec{\text{c}}.\vec{\text{a}}$
  3. $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
  4. $\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
Answer
  1. $\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
Solution:
we have
$\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|\big|\vec{\text{c}}\big|\big|\cos\theta\big|$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|\big|\vec{\text{c}}\big|$ (If $\theta=0^\circ\text{or}180^\circ,$ i.e. vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{c}}$ are parallel)
$=\big|\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\alpha\big)\big|\big|\vec{\text{c}}\big|$
$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ (If $\alpha=90^\circ,$ i. e. vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular)
$\therefore\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ (If vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are perpendicular to each other)
Thus, the relation $\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big|=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\big|\vec{\text{c}}\big|$ holds good if $\vec{\text{a}}.\vec{\text{b}}=0,\vec{\text{b}}.\vec{\text{c}}=0$ and $\vec{\text{c}}.\vec{\text{a}}=0.$
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Question 91 Mark
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-coplanar vectors, then $\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$ is equal to:
  1. 0
  2. 2
  3. 1
  4. None of these
Answer
  1. 0
Solution:
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 tiple 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 in cyclic order of vectors changes the sign of the scalar triple product)
$=1-1$
$=0$
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Question 101 Mark
For any three vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ the expression $\big(\vec{\text{a}}-\vec{\text{b}}\big).\big\{\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big\}$ equals:
  1. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  2. $2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  3. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}]}^2$
  4. None of these
Answer
  1. None of these
Solution:
We have
$\big(\vec{\text{a}}-\vec{\text{b}}\big).\big[\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big]$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big[\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\vec{\text{c}}-\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\vec{\text{a}}\big]$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{c}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times{\vec{\text{c}}}-0-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{b}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{a}}\big)\\+\vec{\text{b}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{a}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)-\vec{\text{b}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-0-0+0+0-\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$ $\big(\therefore\big[\vec{\text{b}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\big]=\big[\vec{\text{b}}\vec{\text{b}}\vec{\text{a}}\big]=0\big)$
$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]\big)$
$=0$
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Question 111 Mark
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}} $ are three non-coplanar mutually perpendicular unit vectors, then $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big],$ is:
  1. $\pm 1$
  2. $0$
  3. $-2$
  4. $2$
Answer
  1. $\pm1$
Solution:
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|\big|\vec{\text{c}}\big|\cos0^\circ$ or $\big|\vec{\text{a}}\times{\vec{\text{b}}}\big|\big|\vec{\text{c}}\big|\cos180^\circ$ $\big(\therefore\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are perpendicular to each other$)$
$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$ or $-\big|\vec{\text{a}}\times\vec{\text{b}}\big|$  $\big(\therefore\big|\vec{\text{c}}\big|=1,\cos0^\circ=1\text{ and }\cos180^\circ=-1\big)$  $$
$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin90^\circ$ or $-\big|\text{a}\big|\big|\vec{\text{b}}\big|\sin90$  $\big(\therefore\vec{\text{a}} \text{ is perpendicular to }\vec{\text{b}})$ 
$=1 \text{ or }-1$ $\big(\therefore\big|\vec{\text{a}}\big|=1 \text{ and }\big|\vec{\text{b}}\big|=1\big)$ $$
$=\pm1$
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MCQ 121 Mark
If the vectors $4\hat{\text{i}}+11\hat{\text{j}}+\text{m}\hat{\text{k}},7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$ and $\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ are coplanar, then $m =$
  • A
    $0$
  • B
    $38$
  • C
    $-10$
  • ✓
    $10$
Answer
Correct option: D.
$10$
Let
$\vec{\text{a}}=4\hat{\text{i}}+11\hat{\text{j}}+{\text{m}}\hat{\text{k}}$
$\vec{\text{b}}=7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are coplanar if their scalar triple product is zero, i.e. _{$\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big]=0$}
$\Rightarrow\begin{vmatrix}4&11&\text{m}\\7&2&6\\1&5&4 \end{vmatrix}=0$
$\Rightarrow 4(8-30)-11(28-6)+\text{m}(35-2)=0$
$\Rightarrow-88-242+33\text{m}=0$
$\Rightarrow33\text{m}=330$
$\therefore\text{m}=10$
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Question 131 Mark
If $\vec{\text{r}}.\vec{\text{a}}=\vec{\text{r}}.\vec{\text{b}}=\vec{\text{r}}.\vec{\text{c}}=0$ for some non-zero vector $\vec{\text{r}},$ then the value of $\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big],$ is:
  1. 2
  2. 3
  3. 0
  4. None of these
Answer
  1. 0
Solution:
If $\vec{\text{r}}.\vec{\text{a}}=0$ for some non-zero vector $\vec{\text{r}},$ then either $\vec{\text{a}}$ is a zero - vector or it is perpendicular to $\vec{\text{r}}.$
If one of $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ is zero, then $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
If all $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are non - zero, then they must be coplanar as they are perpendicular to vector $\vec{\text{r}}.$
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=0$
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Question 141 Mark
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=$
  1. $\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
  2. $\big|\vec{\text{a}}+\vec{\text{b}}\big|^2$
  3. $\big|\vec{\text{a}}\big|^2+\big|\vec{\text{b}}\big|^2$
  4. $2\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
Answer
  1. $\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
Solution:
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\theta\big)^2+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta\big)^2$
$=\big|\vec{\text{a}}\big|^2\big|{\text{b}}\big|^2\sin^2\theta+\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\cos^2\theta$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\sin^2\theta+\cos^2\theta\big)$
$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2$
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Question 151 Mark
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)=$
  1. $0$
  2. $-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  3. $2\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
  4. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
Answer
  1. $\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
Solution:
We have
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}.\big[\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{a}}+\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{b}}+\big(\vec{\text{b}}+\vec{\text{c}}\big)\times\vec{\text{c}}\big]$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}+\vec{\text{b}}\times\vec{\text{b}}+\vec{\text{c}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{c}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}+0+\vec{\text{c}}\times\vec{\text{b}}+\vec{\text{b}}\times\vec{\text{c}}+0$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}-\vec{\text{b}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$
$=\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{b}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{a}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)+\vec{\text{b}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$
$=0+0+0+\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$
$=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$
$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$
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