JEE MAIN Vector Previous Year Questions (PYQs) – Page 15 of 17
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Let $A(3,0,-1),; B(2,10,6)$ and $C(1,2,1)$ be the vertices of a triangle and $M$ be the midpoint of $AC$. If $G$ divides $BM$ in the ratio $2:1$, then $\cos(\angle GOA)$ ($O$ being the origin) is equal to:
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Solution
In a triangle ABC, if $\left| {\overrightarrow {BC} } \right| = 3$, $\left| {\overrightarrow {CA} } \right| = 5$ and $\left| {\overrightarrow {BA} } \right| = 7$, then the projection of the vector $\overrightarrow {BA} $ on $\overrightarrow {BC} $ is equal to :
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Solution
Let $\vec a$ and $\vec b$ be two vectors. Let $|\vec a|=1$, $|\vec b|=4$ and $\vec a\cdot\vec b=2$. If $\vec c=(2\,\vec a\times\vec b)-3\vec b$, then the value of $\vec b\cdot\vec c$ is:
(A) $-48$
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Solution
If the shortest distance between the lines
\[
\frac{x-\lambda}{2}=\frac{y-2}{1}=\frac{z-1}{1}
\quad\text{and}\quad
\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}
\]
is $1$, then the sum of all possible values of $\lambda$ is:
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Solution
Consider a $\triangle ABC$ where $A(1,3,2)$, $B(-2,8,0)$ and $C(3,6,7)$.
If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$, then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:
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Solution
Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive $y$- and $z$-axes, respectively, is half of the angle that this line makes with the positive $x$-axis. Then the sum of all possible values of the angle $\beta$ is
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Solution
Let $a_1,a_2,a_3,\ldots$ be an A.P. with $a_6=2$. Then the common difference of this A.P., which maximises the product $a_1a_4a_5$, is:
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Solution
Let a line pass through two distinct points $P(-2,-1,3)$ and $Q$, and be parallel to the vector $3\hat i+2\hat j+2\hat k$. If the distance of the point $Q$ from the point $R(1,3,3)$ is $5$, then the square of the area of $\triangle PQR$ is equal to:
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Solution
Let $\triangle ABC$ be a triangle whose circumcentre is at $P$.
If the position vectors of $A, B, C$ and $P$ are $\vec a, \vec b, \vec c$ and $\dfrac{\vec a + \vec b + \vec c}{4}$ respectively,
then the position vector of the orthocentre of this triangle is:
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Solution
Let $\vec a=\hat i+\hat j+\sqrt{2}\,\hat k$, $\vec b=b_1\hat i+b_2\hat j+\sqrt{2}\,\hat k$, $\vec c=5\hat i+\hat j+\sqrt{2}\,\hat k$ be three vectors such that the projection vector of $\vec b$ on $\vec a$ is $\vec a$. If $\vec a+\vec b$ is perpendicular to $\vec c$, then $|\vec b|$ is equal to:
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Solution
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