JEE MAIN Vector Previous Year Questions (PYQs) – Page 6 of 17

Let O be the origin and the position vector of the point P be $ - \widehat i - 2\widehat j + 3\widehat k$. If the position vectors of the points A, B and C are $ - 2\widehat i + \widehat j - 3\widehat k,2\widehat i + 4\widehat j - 2\widehat k$ and $ - 4\widehat i + 2\widehat j - \widehat k$ respectively, then the projection of the vector $\overrightarrow {OP} $ on a vector perpendicular to the vectors $\overrightarrow {AB} $ and $\overrightarrow {AC} $ is :

Let $\hat{a}$ and $\hat{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{4}$. If $\theta$ is the angle between the vectors $(\hat{a}+\hat{b})$ and $(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))$, then the value of $164 \,\cos ^{2} \theta$ is equal to :

Let $O$ be the origin and the position vectors of $A$ and $B$ be $2\hat i+2\hat j+\hat k$ and $2\hat i+4\hat j+4\hat k$ respectively. If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$, then the length of $OC$ is:

If the image of the point $(4,4,3)$ in the line $\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-1}{3}$ is $(\alpha,\beta,\gamma)$, then $\alpha+\beta+\gamma$ is equal to

The arc $PQ$ of a circle subtends a right angle at its centre $O$. The midpoint of the arc $PQ$ is $R$. If $\overrightarrow{OP}=\vec{u}$, $\overrightarrow{OR}=\vec{v}$ and $\overrightarrow{OQ}=\alpha\vec{u}+\beta\vec{v}$, then $\alpha,\ \beta^{2}$ are the roots of the equation:

Let the number $(22)^{2022} + (2022)^{22}$ leave the remainder $\alpha$ when divided by $3$ and $\beta$ when divided by $7$. Then $(\alpha^2 + \beta^2)$ is equal to:

Let a unit vector $\hat{\mathbf u}=x\hat i+y\hat j+z\hat k$ make angles $\dfrac{\pi}{2},\ \dfrac{\pi}{3}$ and $\dfrac{2\pi}{3}$ with the vectors $\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat k$, $\dfrac{1}{\sqrt2}\hat j+\dfrac{1}{\sqrt2}\hat k$ and $\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat j$ respectively. If $\vec v=\dfrac{1}{\sqrt2}\hat i+\dfrac{1}{\sqrt2}\hat j+\dfrac{1}{\sqrt2}\hat k$, then $|\hat{\mathbf u}-\vec v|^{2}$ is equal to:

If vectors $\overrightarrow {{a_1}} = x\widehat i - \widehat j + \widehat k$ and $\overrightarrow {{a_2}} = \widehat i + y\widehat j + z\widehat k$ are collinear, then a possible unit vector parallel to the vector $x\widehat i + y\widehat j + z\widehat k$ is :

Let a vector $\alpha \widehat i + \beta \widehat j$ be obtained by rotating the vector $\sqrt 3 \widehat i + \widehat j$ by an angle 45$^\circ$ about the origin in counterclockwise direction in the first quadrant. Then the area of triangle having vertices ($\alpha$, $\beta$), (0, $\beta$) and (0, 0) is equal to :

Let $\vec a=2\hat i+\hat j-\hat k,\quad \vec b=\big((\vec a\times(\hat i+\hat j))\times\hat i\big)\times\hat i.$ Then the square of the projection of $\vec a$ on $\vec b$ is: