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Volume of parallelopiped, $V = \left ( \vec a \times \vec b \right ). \vec c$

Here $\vec a = -12i+\lambda k$

$\vec b = 3j - k$

and, $\vec c = 2i+j-15k$

$\vec a \times \vec b = \begin{bmatrix} i & j & k \\ -12 & 0 & \lambda \\ 0& 3 & -1 \end{bmatrix}$

$\Rightarrow \vec a \times \vec b = -3\lambda i -12j -36k$

$V = \left ( -3\lambda i -12j -36k \right ).(2i+j-15k) = -6\lambda -12 +540 = -6\lambda +528$

Given that V is 546, put it in the above equation.

$546 = -6\lambda +528$

$\Rightarrow \lambda =-3$

Option (4) is correct

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Posted by

Abhishek Sahu

Let $\vec{P}$ is the p.v of the orthocentre and $\vec{g}$ is the p.v of the centroid of the triangle ABC, where circumcentre is the origin. If $\vec{P}= K\vec{g}\: then \: K$

Option 1)
3

Option 2)
2

Option 3)
$\frac{1}{3}$

Option 4)
$\frac{2}{3}$

Use the concept of

Position vector -

Let O be a fixed origin, then position vector of P is $\overrightarrow{OP}$

- wherein

Centroid is intersection is $\frac{x_{1}+x_{2}+x_{3}}{3} , \frac{y_{1}+y_{2}+y_{3}}{3}$

orthocentre: intersection of altitudes.

so that in a triangle ABC the orthocentre H, centroid G and circumcentre M are collinear ang G divides HM internally in the ration 2:1

$\therefore g=\frac{2\times M+1\times P}{3}$

$\Rightarrow p=3g$

$\therefore k=3$ $\left [ \because M=\left ( 0,0,0 \right ) \right ]$

Option 1)

3

Correct option

Option 2)

2

Incorrect option

Option 3)

$\frac{1}{3}$

Incorrect option

Option 4)

$\frac{2}{3}$

Incorrect option

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Posted by

Vakul

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The points with position vectors

$\vec{3i}-\vec{4j}-\vec{4k}, \vec{2i}-\vec{j}+\vec{k}\: and\: \vec{i}-\vec{3j}-\vec{5k}$

Option 1)
An Equilateral Triangle

Option 2)
An Isosceles Triangle

Option 3)
A Right Angle Triangle

Option 4)
None of these

As we learnt

Scalar Product of two vectors -

$\vec{a}.\vec{b}> 0 \:an\: acute\: angle$

$\vec{a}.\vec{b}< 0 \:an\: obtuse\: angle$

$\vec{a}.\vec{b}= 0 \:a\:right\: angle$

- wherein

$\Theta$ is the angle between the vectors $\vec{a}\:and\:\vec{b}$

$\vec{A}=3i-4j-4k$

$\vec{B}=2i-j+k$

$\vec{C}=i-3j-5k$

$\underset{AB}{\rightarrow}= -i+3j+5k$

$\underset{AC}{\rightarrow}= -2i+j-k$

$\underset{AB}{\rightarrow}.\underset{AC}{\rightarrow}=2+3-5=0$

So that it is right angle triangle

Option 1)

An Equilateral Triangle

Incorrect option

Option 2)

An Isosceles Triangle

Incorrect option

Option 3)

A Right Angle Triangle

Correct option

Option 4)

None of these

Incorrect option

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Posted by

Plabita

If $\left | \vec{a} \right |=\left | \vec{b} \right |, then (\vec{a}+\vec{b}).(\vec{a}-\vec{b}) \:is$

Option 1)
Positive

Option 2)
Negative

Option 3)
Zero

Option 4)
None

As we learnt

Scalar Product of two vectors -

$\vec{a}.\vec{b}> 0 \:an\: acute\: angle$

$\vec{a}.\vec{b}< 0 \:an\: obtuse\: angle$

$\vec{a}.\vec{b}= 0 \:a\:right\: angle$

- wherein

$\Theta$ is the angle between the vectors $\vec{a}\:and\:\vec{b}$

$\left | \vec{a} \right |=\left | \vec{b} \right |$

$\left ( \vec{a}+\vec{b} \right ).\left ( \vec{a}-\vec{b} \right )=\vec{a}.\vec{a}-\vec{a}\vec{b}+\vec{b}.\vec{a}-\vec{b}.\vec{b}$

$\left | \vec{a} \right |^{2}-\left | \vec{b} \right |^{2}$ $\left [ \because \left | a \right | \right=\left | b \right | ]$

= 0

Option 1)

Positive

Incorrect Option

Option 2)

Negative

Incorrect Option

Option 3)

Zero

Correct Option

Option 4)

None

Incorrect Option

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Posted by

Aadil

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If vector $2\vec{i}+3\vec{j}-2\vec{k}$ and $\vec{i}+\vec{2j}+\vec{k}$ represents the adjacent sides of any parallelogram then the lenght of diagonal of parallelogram are

Option 1)
$\sqrt{35},\sqrt{35}$

Option 2)
$\sqrt{35}, \sqrt{11}$

Option 3)
$\sqrt{25},\sqrt{11}$

Option 4)
None of these

Magnitude of a Vector -

The length of the directed line segment $\overrightarrow{AB}$ is called its magnitude.

- wherein

It is denoted by $\mid \overrightarrow{AB\mid }$

Diagonal = 2i+3j-2k+i+2j+k

$\vec{d}=3i+5j-k$

$\left | \vec{d} \right |=\sqrt{3^{2}+5^{2}+1^{2}}$

$=\sqrt{9+25+1}$

$=\sqrt{35}$

Similarly

$\vec{d}=i+j-3k$

$\left | \vec{d} \right |=\sqrt{1+1+9}$

$=\sqrt{11}$

Option 1)

$\sqrt{35},\sqrt{35}$

Incorrect Option

Option 2)

$\sqrt{35}, \sqrt{11}$

Correct option

Option 3)

$\sqrt{25},\sqrt{11}$

Incorrect Option

Option 4)

None of these

Incorrect Option

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Posted by

divya.saini

If vector $\vec{a}+\vec{b}$ is perpendicular to $\vec{b}$ and $\vec{2b}+\vec{a}$ is perpendicular to $\vec{a}$ , then

Option 1)
$\left | \vec{a} \right |=\sqrt{2}\left | \vec{b} \right |$

Option 2)
$\left | \vec{a} \right |={2}\left | \vec{b} \right |$

Option 3)
$\left | \vec{b} \right |=\sqrt{2}\left | \vec{a} \right |$

Option 4)
$\left | \vec{a} \right |=\left | \vec{b} \right |$

Use concept ID

Scalar Product of two vectors -

$\vec{a}.\vec{b}> 0 \:an\: acute\: angle$

$\vec{a}.\vec{b}< 0 \:an\: obtuse\: angle$

$\vec{a}.\vec{b}= 0 \:a\:right\: angle$

- wherein

$\Theta$ is the angle between the vectors $\vec{a}\:and\:\vec{b}$

$\left ( \vec{a} \right+\vec{b} ).\vec{b}=0 \Rightarrow \vec{a}.\vec{b}+\vec{b}.\vec{b}=0$

and $\left ( 2\vec{b}+\vec{a} \right).\vec{a}=0\Rightarrow 2\vec{b}.\vec{a}+\vec{a}.\vec{a}=0$

$\therefore \vec{a}.\vec{b}+\left | b \right |^{2}=2\vec{a}.\vec{b}+\left | a \right |^{2}$

= $\left | \vec{b} \right |^{2}-\left | \vec{a} \right |^{2}=\vec{a}.\vec{b}$

$\therefore \left | b \right |^{2}-\left | a \right |^{2}+\left | b \right |^{2}=0$

$\left | a \right |^{2}=2\left | b \right |^{2}$

$\left | \vec{a} \right |=\sqrt{2}\left | \vec{b} \right |$

Option 1)

$\left | \vec{a} \right |=\sqrt{2}\left | \vec{b} \right |$

Correct option

Option 2)

$\left | \vec{a} \right |={2}\left | \vec{b} \right |$

Incorrect option

Option 3)

$\left | \vec{b} \right |=\sqrt{2}\left | \vec{a} \right |$

Incorrect option

Option 4)

$\left | \vec{a} \right |=\left | \vec{b} \right |$

Incorrect option

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Posted by

prateek

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If the vector $\vec{b}$ is collinear with the vector $\vec{a} =(2\sqrt{2},-1,4) \:and \:\left | \vec{b} \right |=10, \:then$

Option 1)
$\vec{a}+\vec{b}=0$

Option 2)
$\vec{a}+\vec{2b}=0$

Option 3)
$\vec{2a}\ ^+_-\ \vec{b}=0$

Option 4)
None

Use the concept

Magnitude of a Vector -

The length of the directed line segment $\overrightarrow{AB}$ is called its magnitude.

- wherein

It is denoted by $\mid \overrightarrow{AB\mid }$

given that

$\vec{b}=\lambda \left ( 2\sqrt{2}i-j+4k \right )$

$\left | \vec{b} \right |^{2}=8\lambda ^{2}+\lambda ^{2}+16\lambda ^{2}=25\lambda ^{2}$

$=\frac{100}{25}=\lambda ^{2}=4$

$\therefore \lambda =\pm 2$

$\vec{b}=4\sqrt{2}i-2j+8k$

or

$\vec{b}= -2\sqrt{2}i+2i-8k$

$\vec{b}= \pm \underset{2a}{\rightarrow}$

$\therefore \vec{b}\pm 2\vec{a}=0$

Option 1)

$\vec{a}+\vec{b}=0$

Incorrect option

Option 2)

$\vec{a}+\vec{2b}=0$

Incorrect option

Option 3)

$\vec{2a}\ ^+_-\ \vec{b}=0$

Correct option

Option 4)

None

Incorrect option

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Posted by

prateek

If position vectors of A, B, C, D are respectively $\vec{2i}+\vec{3j}+\vec{5k}, \vec{i}+\vec{2j}+\vec{3k}, -\vec{5i}+\vec{4j}-\vec{2k}$ and $\vec{i}+\vec{10j}+\vec{10k},$ then

Option 1)
$AB\parallel CD$

Option 2)
$DC\parallel AD$

Option 3)
A,B,C are collinear

Option 4)
B,C,D are collinear

As we learnt

Position vector -

Let O be a fixed origin, then position vector of P is $\overrightarrow{OP}$

- wherein

Since $\underset{AB}{\rightarrow}=\left ( p.v \right )\vec{B}-\left ( p.v \right )\vec{A}$

So that $\underset{AB}{\rightarrow}=\left ( i+2j+3k \right )-\left ( 2i=3j+5k \right )$

$= -i-j-2k$

Similarly $\underset{CD}{\rightarrow}=6i+6j+12k$

$= 6\left ( i+j+2k \right )$

So that ratios are equal. so they are parellal

$\underset{AB}{\rightarrow}\left | \right |\underset{CD}{\rightarrow}$

Option 1)

$AB\parallel CD$

Correct option

Option 2)

$DC\parallel AD$

Incorrect Option

Option 3)

A,B,C are collinear

Incorrect Option

Option 4)

B,C,D are collinear

Incorrect Option

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Posted by

divya.saini

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If position vectors of A, B, C, D are respectively $\vec{2i}+\vec{3j}+\vec{5k}, \vec{i}+\vec{2j}+\vec{3k}, -\vec{5i}+\vec{4j}-\vec{2k}$ and $\vec{i}+\vec{10j}+\vec{10k},$ then

Option 1)
$AB\parallel CD$

Option 2)
$DC\parallel AD$

Option 3)
$A, B, C,$ are collinear

Option 4)
$B,C,D$ are collinear

As we learnt

Position vector -

Let O be a fixed origin, then position vector of P is $\overrightarrow{OP}$

- wherein

$A= 2i+3j+5k$

$B= i+2j+3k$

$C=-5i+4j-2k$

$D=i+10j+10k$

$A\vec{B}= -i-j-2k= -\left ( i+j+k \right )$

$C\vec{D}= 6i+6j+12k$

$=6\left ( i+j+2k \right )$

So $\overrightarrow{AB}=\lambda \overrightarrow{CD}$

So $\overrightarrow{AB}\left | \right | \overrightarrow{CD}$

So

Option 1)

$AB\parallel CD$

Correct option

Option 2)

$DC\parallel AD$

Incorrect option

Option 3)

$A, B, C,$ are collinear

Incorrect option

Option 4)

$B,C,D$ are collinear

Incorrect option

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Posted by

Aadil

Angle between the vectors $\vec{2i}+\vec{6j}+\vec{3k} \:and \:\vec{12i}-\vec{4j}+\vec{3k} \:is$

Option 1)
$cos^-(\frac{1}{10})$

Option 2)
$cos^-(\frac{9}{11})$

Option 3)
$cos^-(\frac{9}{91})$

Option 4)
$cos^-(\frac{1}{9})$

As we learnt

Angle between vector a and vector b -

$\cos \Theta =\frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}$

- wherein

Here $0\leq \Theta \leq \pi$ ??????

$\cos \Theta =\frac{\vec{a.\vec{b}}}{\left | \vec{a} \right |\left | \vec{b} \right |}$

= $\frac{\left ( 2i+6j+3\vec{k} \right )}{\sqrt{2^{2}+6^{2}+3^{2}}}.\frac{\left ( 12i-4j+3k \right )}{\sqrt{12^{2}+4^{2}+3^{2}}}$

$=\frac{24-24+9}{\sqrt{49}\times \sqrt{169}}$

$=\frac{9}{7\times 13}$ $=\frac{9}{7\times 13}= \frac{9}{91}$

$\Theta =\cos ^{-1}\left ( \frac{9}{91} \right )$

Option 1)

$cos^-(\frac{1}{10})$

Incorrect Option

Option 2)

$cos^-(\frac{9}{11})$

Incorrect Option

Option 3)

$cos^-(\frac{9}{91})$

Correct Option

Option 4)

$cos^-(\frac{1}{9})$

Incorrect Option

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Posted by

Plabita

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

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