If a=2 hat{i}+3 hat{j}+hat{k}, b=hat{i}-3 hat | vedclass

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

Question

(A-(IV), B-(III), C-(II), D-(I)) Given: $a=2 \hat{i}+3 \hat{j}+\hat{k}$, $b=\hat{i}-3 \hat{j}-5 \hat{k}$, $c=3 \hat{i}-4 \hat{k}$.$A$. $a-b = (2-1)\ha

Vedclass · Accepted Answer

Vedclass · Answer

(A-(IV), B-(III), C-(II), D-(I)) Given: $a=2 \hat{i}+3 \hat{j}+\hat{k}$, $b=\hat{i}-3 \hat{j}-5 \hat{k}$, $c=3 \hat{i}-4 \hat{k}$.$A$. $a-b = (2-1)\hat{i} + (3-(-3))\hat{j} + (1-(-5))\hat{k} = \hat{i} + 6\hat{j} + 6\hat{k}$.The vector opposite to $a-b$ is $-(a-b) = -\hat{i} - 6\hat{j} - 6\hat{k}$.Magnitude is $\sqrt{(-1)^2 + (-6)^2 + (-6)^2} = \sqrt{1+36+36} = \sqrt{73}$.Unit vector is $\frac{-\hat{i}-6\hat{j}-6\hat{k}}{\sqrt{73}} = -\frac{\hat{i}}{\sqrt{73}} - \frac{6\hat{j}}{\sqrt{73}} - \frac{6\hat{k}}{\sqrt{73}}$. Matches $(iv)$.$B$. In $	riangle ABC$, $\vec{AB} + \vec{BC} + \vec{CA} = 0$.So, $\vec{CA} = -(\vec{AB} + \vec{BC}) = -(a+b) = -(2\hat{i}+3\hat{j}+\hat{k} + \hat{i}-3\hat{j}-5\hat{k}) = -(3\hat{i}-4\hat{k}) = -3\hat{i}+4\hat{k}$. Matches $(iii)$.$C$. Centroid $G = \frac{a+b+c}{3} = \frac{(2\hat{i}+3\hat{j}+\hat{k}) + (\hat{i}-3\hat{j}-5\hat{k}) + (3\hat{i}-4\hat{k})}{3} = \frac{6\hat{i} + 0\hat{j} - 8\hat{k}}{3} = 2\hat{i} - \frac{8}{3}\hat{k}$. Matches $(ii)$.$D$. $d$ is parallel to $a$, so $d = k a = k(2\hat{i}+3\hat{j}+\hat{k})$.Magnitude $|d| = |k|\sqrt{2^2+3^2+1^2} = |k|\sqrt{14}$.Given $|d| = 2\sqrt{14}$, so $|k|=2$. Taking $k=2$, $d = 4\hat{i}+6\hat{j}+2\hat{k}$.Then $b+d = (\hat{i}-3\hat{j}-5\hat{k}) + (4\hat{i}+6\hat{j}+2\hat{k}) = 5\hat{i}+3\hat{j}-3\hat{k}$. Matches $(i)$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$, then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then, its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$, then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

Similar Questions

$A, B, P, Q, R$ are five points in a plane. If the forces acting at point $A$ are $\overline{AP}, \overline{AQ}, \overline{AR}$ and the forces acting at point $B$ are $\overline{PB}, \overline{QB}, \overline{RB}$,find the resultant of all these forces.

For what values of $\lambda$ are $\vec{a}$ and $\vec{c}$ unit collinear vectors,and given $|\vec{b}| = 6$,if $\vec{b} - 3\vec{c} = \lambda \vec{a}$,then $\lambda = ......$

Which of the following is not always true?

If points $D, E, F$ divide the sides $BC, CA, AB$ of $\triangle ABC$ in the ratios $1:4, 3:2, 3:7$ respectively,and point $K$ divides $AB$ in some ratio,then $(\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}) : \overrightarrow{CK} = ......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module