Match List I with List II and select the corr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) $(P)$ Given $[\vec{a} \vec{b} \vec{c}] = 2$. The volume of the parallelepiped determined by $2(\vec{a} 	imes \vec{b}), 3(\vec{b} 	imes \vec{c})$

Vedclass · Accepted Answer

$3 \quad 4 \quad 1 \quad 2$

Vedclass · Answer

(C) $(P)$ Given $[\vec{a} \vec{b} \vec{c}] = 2$. The volume of the parallelepiped determined by $2(\vec{a} 	imes \vec{b}), 3(\vec{b} 	imes \vec{c})$ and $(\vec{c} 	imes \vec{a})$ is given by the scalar triple product $|2(\vec{a} 	imes \vec{b}) \cdot (3(\vec{b} 	imes \vec{c}) 	imes (\vec{c} 	imes \vec{a}))|$.Using the property $(\vec{b} 	imes \vec{c}) 	imes (\vec{c} 	imes \vec{a}) = [\vec{b} \vec{c} \vec{a}] \vec{c} = [\vec{a} \vec{b} \vec{c}] \vec{c}$,we get $6[\vec{a} \vec{b} \vec{c}]^2 = 6(2)^2 = 24$.Thus,$P ightarrow 3$.$(Q)$ Given $[\vec{a} \vec{b} \vec{c}] = 5$. The volume is $[3(\vec{a}+\vec{b}) \quad (\vec{b}+\vec{c}) \quad 2(\vec{c}+\vec{a})] = 6 [(\vec{a}+\vec{b}) \cdot ((\vec{b}+\vec{c}) 	imes (\vec{c}+\vec{a}))]$.Since $(\vec{b}+\vec{c}) 	imes (\vec{c}+\vec{a}) = \vec{b} 	imes \vec{c} + \vec{b} 	imes \vec{a} + \vec{c} 	imes \vec{c} + \vec{c} 	imes \vec{a} = \vec{b} 	imes \vec{c} + \vec{b} 	imes \vec{a} + \vec{c} 	imes \vec{a}$,the triple product becomes $6 	imes 2 [\vec{a} \vec{b} \vec{c}] = 12 	imes 5 = 60$.Thus,$Q ightarrow 4$.$(R)$ Area of triangle $= \frac{1}{2} |\vec{a} 	imes \vec{b}| = 20$,so $|\vec{a} 	imes \vec{b}| = 40$. The new area is $\frac{1}{2} |(2\vec{a}+3\vec{b}) 	imes (\vec{a}-\vec{b})| = \frac{1}{2} |-2(\vec{a} 	imes \vec{b}) + 3(\vec{b} 	imes \vec{a})| = \frac{1}{2} |-2(\vec{a} 	imes \vec{b}) - 3(\vec{a} 	imes \vec{b})| = \frac{5}{2} |\vec{a} 	imes \vec{b}| = \frac{5}{2} 	imes 40 = 100$.Thus,$R ightarrow 1$.$(S)$ Area of parallelogram $= |\vec{a} 	imes \vec{b}| = 30$. The new area is $|(\vec{a}+\vec{b}) 	imes \vec{a}| = |\vec{a} 	imes \vec{a} + \vec{b} 	imes \vec{a}| = |\vec{b} 	imes \vec{a}| = |\vec{a} 	imes \vec{b}| = 30$.Thus,$S ightarrow 2$.

List $I$	List $II$
$P$. Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $2$. Then the volume of the parallelepiped determined by vectors $2(\vec{a} \times \vec{b}), 3(\vec{b} \times \vec{c})$ and $(\vec{c} \times \vec{a})$ is	$1$. $100$
$Q$. Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $5$. Then the volume of the parallelepiped determined by vectors $3(\vec{a}+\vec{b}), (\vec{b}+\vec{c})$ and $2(\vec{c}+\vec{a})$ is	$2$. $30$
$R$. Area of a triangle with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $20$. Then the area of the triangle with adjacent sides determined by vectors $(2\vec{a}+3\vec{b})$ and $(\vec{a}-\vec{b})$ is	$3$. $24$
$S$. Area of a parallelogram with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is $30$. Then the area of the parallelogram with adjacent sides determined by vectors $(\vec{a}+\vec{b})$ and $\vec{a}$ is	$4$. $60$

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

Similar Questions

Let $\overrightarrow{OA}=2 \hat{i}-3 \hat{j}+\hat{k}$,$\overrightarrow{OB}=\hat{i}-4 \hat{j}-3 \hat{k}$,and $\overrightarrow{OC}=-3 \hat{i}+\hat{j}+2 \hat{k}$ be the position vectors of three points $A$,$B$,and $C$ respectively. If $G$ is the centroid of triangle $ABC$,then $BC^2+CA^2+AB^2+9(OG)^2=$

If $ABCDEF$ is a regular hexagon,and $\vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF} = k \vec{AD}$,then $k = \dots$

If $b$ and $c$ are non-collinear vectors, $|c| \neq 0$, $a \times(b \times c)+(a \cdot b) b=(4-2 \beta-\sin \alpha) b+\left(\beta^2-1\right) c$ and $(c \cdot c) a=c$, then the scalars $\alpha$ and $\beta$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module