Scalar Triple Product MCQ Quiz in मल्याळम - Objective Question with Answer for Scalar Triple Product - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Scaler triple product of the vectors:

The scalar triple product of the vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) is given by:

Coplanar vectors:

Three vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) are said to be coplanar if the scalar triple product \({\bf{\bar A}} \cdot \left[ {{\bf{\bar B}} \times {\bf{\bar C}}} \right] = 0.\)

Calculation:

\(\left| {\begin{array}{*{20}{c}} a&a&c\\ 1&0&1\\ c&c&b \end{array}} \right| = 0 \)       

⇒ a(-c) - a(b-c) + c(c) = 0 

⇒ c² = ab

Hence c is the geometric mean of a and b.

Concept:

Scalar Triple Product: 

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\), \(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) and \(\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\) then their scalar triple product is defined as \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

Calculation:

Given: \(\vec a = 2\hat i + 3\hat j + 4\hat k\;,\;\vec b = 3\hat i + 4\hat j + 6\hat k\;and\;\vec c = 2\hat i + 4\hat j + \hat k\)

As we know that, the scalar triple product of three vectors is \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{2}}&{{3}}&{{4}}\\ {{3}}&{{4}}&{{6}}\\ {{2}}&{{4}}&{{1}} \end{array}} \right|\)

⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = 2 \times (4 - 24) - 3 \times (3 - 12) + 4 \times (12 - 8) = 3\)

Hence, the correct option is 2.

Concept:

Scalar Triple Product: 

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\), \(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) and \(\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\) then their scalar triple product is defined as \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

Calculation:

Given: \(\vec a = \hat i + \hat j + \hat k\;,\;\vec b = 7\hat i + 2\hat j + 3\hat k\;and\;\vec c = \hat i + \hat j + \hat k\)

As we know that, the scalar triple product of three vectors is \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{1}}&{{1}}&{{1}}\\ {{7}}&{{2}}&{{3}}\\ {{1}}&{{1}}&{{1}} \end{array}} \right|\)

⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = 1 \times (2 - 3) - 1 \times (7 - 3) + 1 \times (7 - 2) = 0\)

Hence, the correct option is 4.

Concept:

If \(\rm \vec{a}=a_{1}\hat i+a_{2}\hat j+a_{3}\hat k\), \(\rm \vec{b}=b_{1}\vec i+b_{2}\vec j+b_{3}\vec k\) and \(\rm \vec{c}=c_{1}\vec i+c_{2}\vec j+c_{3}\vec k\), then \(\rm \vec{a}.( \vec{b}\times \vec{c})= \begin{vmatrix} a_{1} &a_{2} &a_{3}\\ b_{1} & b_{2} &b_{3} \\ c_{1} & c_{2} &c_{3} \end{vmatrix}\).

Note: \(\rm \vec{a}.( \vec{b}\times \vec{c})=\vec{b}.( \vec{c}\times \vec{a})=\vec{c}.( \vec{a}\times \vec{b})\)

Calculation:

Let  \(\rm \vec{a}=a_{1}i+a_{2}j+a_{3}k\), \(\rm \vec{b}=b_{1}i+b_{2}j+b_{3}k\) and \(\rm \vec{c}=c_{1}i+c_{2}j+c_{3}k\),

As we know that, if \(\rm \vec{a}.( \vec{b}\times \vec{c})=\vec{b}.( \vec{c}\times \vec{a})=\vec{c}.( \vec{a}\times \vec{b})\)

Hence, option 2 is correct.

Concept:

Scalar Triple Product: 

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\), \(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) and \(\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\) then their scalar triple product is defined as

\(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

Calculation:

Given: \(\vec a = \hat i + 3\hat j + 4\hat k\;,\;\vec b = 3\hat i + 4\hat j + 2\hat k\;and\;\vec c = 2\hat i + 4\hat j + 5\hat k\)

As we know that, the scalar triple product of three vectors is 

\(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{1}}&{{3}}&{{4}}\\ {{3}}&{{4}}&{{2}}\\ {{2}}&{{4}}&{{5}} \end{array}} \right|\)

\(\Rightarrow 1 \times (20 - 8) - 3 \times (15 - 4) + 4 \times (12 - 8) \\ \Rightarrow-5\)

Hence, the correct option is 3.

Concept:

\(\vec{a}=a_1\hat{i} +a_2\hat{j}+ a_3 \hat{k}\)

\(\rm\vec{b}=b_1\hat{i} +b_2\hat{j}+ b_3 \hat{k}\)

\(\rm \vec{c}=c_1\hat{i} +c_2\hat{j}+ c_3 \hat{k}\)

\(\vec{a}\cdot (\vec{b}\times \vec{c})\) = \(\rm \begin{vmatrix} a_1& a_2 &a_3 \\ b_1& b_2 &b_3 \\ c_1& c_2 &c_3 \end{vmatrix}\)

Calculations:

Given, 

 \(\vec{a}=\hat{i}-\hat{k}\)

\(\vec{b}=x\hat{i}+\hat{j}+(1-x)\hat{k}\)

\(\vec{c}=y\hat{i}+x\hat{j}+(1+x-y)\hat{k}\)

then \(\vec{a}\cdot (\vec{b}\times \vec{c})\) = \(\rm \begin{vmatrix} 1& 0 &-1 \\ x &1 &1-x \\ y &x & 1+x -y \end{vmatrix}\)

⇒\(\vec{a}\cdot (\vec{b}\times \vec{c})\) = [1(1 + x - y) - x( 1 - x)] - 0 - 1(x² - y)

⇒\(\vec{a}\cdot (\vec{b}\times \vec{c})\) = 1 + x - y - x + x²  - x2 + y

⇒\(\vec{a}\cdot (\vec{b}\times \vec{c})\) = 1

Hence,  \(\vec{a}\cdot (\vec{b}\times \vec{c})\) depends on neither on x nor on y.

Concept:

For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:

• Dot Product is defined as \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).
• Cross Product is defined as \(\rm \vec A\times \vec B=\vec n|\vec A||\vec B|\sin \theta\) where \(\rm \vec n\) is the unit vector perpendicular to the plane containing \(\rm \vec A\) and \(\rm \vec B\).

For three vectors \(\rm \vec A\), \(\rm \vec B\) and \(\rm \vec C\):

• Triple Cross Product: is defined as: \(\rm \vec A\times(\vec B\times\vec C)=(\vec A.\vec C)\vec B-(\vec A.\vec B)\vec C\).
• Triple Scalar Product (Box Product): is defined as \(\rm [\vec A\ \vec B\ \vec C]=\vec A.(\vec B\times\vec C)=\begin{vmatrix} \rm a_1 & \rm a_2 & \rm a_3 \\ \rm b_1 & \rm b_2 & \rm b_3 \\\rm c_1 & \rm c_2 & \rm c_3 \end{vmatrix}\).

Volume of a parallelepiped, with vectors \(\rm \vec a\), \(\rm \vec b\) and \(\rm \vec c\) as its sides, is given by the box product of the three vectors.

• Volume = \(\rm [\vec a\ \vec b\ \vec c]\).

Calculation:

The sides of the parallelepiped are:

\(\rm \vec a\) = 2î + 3ĵ + 4k̂

\(\rm \vec b\) = î + αĵ + 2k̂

\(\rm \vec c\) = î + 2ĵ + αk̂

∴ Volume = \(\rm \rm [\vec a\ \vec b\ \vec c]=\begin{vmatrix} 2 & 3 & 4 \\ 1 & \alpha & 2 \\ 1 & 2 & \alpha \end{vmatrix}\) = 15

⇒ 2(α² - 4) + 3(2 - α) + 4(2 - α) = 15

⇒ 2α2 - 8 + 6 - 3α + 8 - 4α = 15

⇒ 2α2 - 7α - 9 = 0

⇒ 2α2 - 9α + 2α - 9 = 0

⇒ α(2α - 9) + (2α - 9) = 0

⇒ (2α - 9)(α + 1) = 0

⇒ 2α - 9 = 0 OR α + 1 = 0

⇒ α = \(\dfrac92\) OR α = -1.

Concept:

Scalar Triple Product: 

If \(\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\), \(\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\) and \(\vec c = {c_1}\hat i + {c_2}\hat j + {c_3}\hat k\) then their scalar triple product is defined as \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

Calculation:

Given: \(\vec a = 2\hat i + 3\hat j + 4\hat k\;,\;\vec b = 3\hat i + 4\hat j + 6\hat k\;and\;\vec c = 2\hat i + 4\hat j + \hat k\)

As we know that, the scalar triple product of three vectors is \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{a_1}}&{{a_2}}&{{a_3}}\\ {{b_1}}&{{b_2}}&{{b_3}}\\ {{c_1}}&{{c_2}}&{{c_3}} \end{array}} \right| = \left[ {a\;b\;c} \right]\)

⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = \left| {\begin{array}{*{20}{c}} {{2}}&{{3}}&{{4}}\\ {{3}}&{{4}}&{{6}}\\ {{2}}&{{4}}&{{1}} \end{array}} \right|\)

⇒ \(\vec a \cdot \left( {\vec b \times \;\vec c} \right) = 2 \times (4 - 24) - 3 \times (3 - 12) + 4 \times (12 - 8) = 3\)

Hence, the correct option is 1.

Concept:

1. The vectors which are parallel to the same plane , or lie on the same plane are called coplanar vectors.

2. Three vectors are coplanar if their scalar triple product is zero. \(⃗{a} \cdot (⃗{b} \times ⃗{c}) = 0 \)

Calculation

Let \(\vec{a} \) = 2i – j + k

\(\vec{b} \) =i + 2j – 3k

\(\vec{c} \)= 3i + a j + 5k

\( ⃗{b} \times ⃗{c} = \begin{vmatrix} \hat{i} &\hat{j}& \hat{k}\\ 1 & 2& -3\\ 3& a& 5\\\end{vmatrix} \)

⇒ \(\hat{i}(10+3a)-\hat{j}(5+9)+\hat{k}(a-6) \)

⇒ \((10+3a)\hat{i}-14\hat{j}+(a-6)\hat{k} \)

⇒ \(\vec {a}\cdot(\vec{b}\times \vec {c})=(2\hat{i}-\hat{j}+\hat{k})\cdot((10+3a)\hat{i}-14\hat{j}+(a-6)\hat{k}) \)

⇒ 2(10 + 3a) - (-14) + (a - 6)

⇒ 20 + 6a + 14 + a - 6

⇒ 7a + 28

Since 7a + 28 = 0

⇒ a = -4

Hence Option(4) is correct.

Concept:

If \(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\) are three vectors, then \(\overrightarrow{a}\cdot(\overrightarrow{b}\times \overrightarrow{c})\) is called the scalar triple product or box product, denoted by [abc]

• [abc] = [bca] = [cab] (It follows cyclic order)

For four vectors, \(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c},\overrightarrow{d}\) quadruple product is given by:

\((\overrightarrow{a}\times\overrightarrow{b})\times (\overrightarrow{c}\times\overrightarrow{d})\) = [abd]c - [abc]d

Calculation:

We have, \(\overrightarrow{p}=\frac{\overrightarrow{b}\times\overrightarrow{c}}{[bca]},\overrightarrow{q}=\frac{\overrightarrow{c}\times\overrightarrow{a}}{[cab]},\overrightarrow{r}=\frac{\overrightarrow{a}\times\overrightarrow{b}}{[abc]}\)

[pqr] = \([\frac{\overrightarrow{b}\times\overrightarrow{c}}{[bca]}\frac{\overrightarrow{c}\times\overrightarrow{a}}{[cab]}\frac{\overrightarrow{a}\times\overrightarrow{b}}{[abc]}]\)

\(\rm[\vec p\ \vec q\ \vec r]=\frac{1}{[\vec a\ \vec b\ \vec c]}\left[(\vec b\times \vec c)\ (\vec c \times \vec a)\ (\vec a \times \vec b)\right]\)

\(\Rightarrow \rm[\vec p \vec q\vec r]=\frac{1}{[\vec a\vec b\vec c]}(\vec b\times \vec c)\left((\vec c \times \vec a)\times(\vec a \times \vec b)\right)\)      ...........(1)

∵ \(\rm (\vec c \times \vec a)\times(\vec a \times \vec b)=\left((\vec c\times \vec a).\vec b\right)\vec a-\left((\vec c \times \vec a).\vec a\right)\vec b\)

\(\Rightarrow \rm (\vec c \times \vec a)\times(\vec a \times \vec b)=[\vec c\ \vec a\ \vec b]\vec a-0\)

\(\Rightarrow \rm (\vec c \times \vec a)\times(\vec a \times \vec b)=[\vec a\ \vec b\ \vec c]\vec a\)      .......(2)

put this in (1), 

\(\Rightarrow \rm[\vec p \vec q\vec r]=\frac{1}{[\vec a\vec b\vec c]}(\vec b\times \vec c).[\vec a\ \vec b\ \vec c]\vec a\)

\(\rm=\frac{[\vec a\ \vec b \ \vec c]}{[\vec a\ \vec b\ \vec c]}(\vec b\times \vec c).\vec a\)

\(\rm =1[\vec b\ \vec c\ \vec a]=[\vec a\ \vec b\ \vec c]\)

The correct answer is Option 1