Vector Algebra Questions (573)

Let $\vec{a} = 2\hat{i}+\hat{j}+\hat{k}$, and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $|\vec{a}+\vec{b}+\vec{c}| = |\vec{a}+\vec{b}-\vec{c}|$ and $\vec{b}\cdot\vec{c}=0$. Consider the following two statements: (A) $|\vec{a}+\lambda\vec{c}| \geq |\vec{a}|$ for all $\lambda \in \mathbb{R}$. (B) $\vec{a}$ and $\vec{c}$ are always parallel. Then:

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

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be $\vec{a}-\vec{b}+\vec{c}$, $\lambda\vec{a}-3\vec{b}+4\vec{c}$, $-\vec{a}+2\vec{b}-3\vec{c}$ and $2\vec{a}-4\vec{b}+6\vec{c}$ respectively. If $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar, then $\lambda$ is:

Let a, b and c be three vectors that \([\vec{a}\;\vec{b}\;\vec{c}] = 2\). If \(\vec{r} = l(\vec{b}\times\vec{c}) + m(\vec{c}\times\vec{a}) + n(\vec{a}\times\vec{b})\) be perpendicular to \(\vec{a}+\vec{b}+\vec{c}\), then the value of \(l+m+n\) is ________.

If $|\vec{a}| = |\vec{b}| = |\vec{c}| = 2$ and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = -1$, then $|\vec{a} \times \vec{b} \times \vec{c} \times \vec{a}|$ is_______.

Sum of all $\alpha$ for which $\hat{i}-2\hat{j}+3\hat{k}$, $2\hat{i}-3\hat{j}+4\hat{k}$, $(\alpha+1)\hat{i}+2\hat{k}$, $9\hat{i}+(\alpha-8)\hat{j}+6\hat{k}$ are coplanar, is equal to

Let $P$ be a point in the plane of the vectors $\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}$ such that $P$ is equidistant from the lines AB and AC. If $|\overrightarrow{AP}|=\dfrac{\sqrt{5}}{2}$, then the area of the triangle ABP is:

Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{31}$, $4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a}\times\vec{b}) = 3(\vec{c}\times\vec{a})$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$, then $\left(\frac{\vec{a}\times\vec{c}}{\vec{a}\cdot\vec{b}}\right)^2$ is equal to _____.

In a trapezium, the vector \(\vec{BC} = \lambda \vec{AD}\). We will then find that \(\vec{p} = \vec{AC} + \vec{BD}\) is collinear with \(\vec{AD}\). If \(\vec{p} = \mu \vec{AD}\), then find the relationship between \(\lambda\) and \(\mu\).

Given that the vectors $\vec{a}, \vec{b}$ and $\vec{c}$ (no two of them are collinear). Further if $(\vec{a} + \vec{b})$ is collinear with $\vec{c}, (\vec{b} + \vec{c})$ is collinear with $\vec{a}$ and $|\vec{a}| = |\vec{b}| = |\vec{c}| = \sqrt{2}$. Then the value of $|\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}|$ is ______.

Let $\vec{u}, \vec{v}, \vec{w}$ be such that $|\vec{u}| = 1$, $|\vec{v}| = 2$, $|\vec{w}| = 3$. If the projection $\vec{v}$ along $\vec{u}$ is equal to that of $\vec{w}$ along $\vec{u}$ and $\vec{v}, \vec{w}$ are perpendicular to each other, then $\frac{|\vec{u} - \vec{v}|^2}{2}$ equals _______.

We have \(|2\vec{a} - \vec{b}|^2 = 25 \Rightarrow 4a^2 + b^2 - 4\vec{a}\cdot\vec{b} = 25\). Given \(|\vec{a}| = 2, |\vec{b}| = 3\), find \(|2\vec{a} + \vec{b}|^2\):

A particle is acted upon by constant forces \(4\hat{i}+\hat{j}-3\hat{k}\) and \(3\hat{i}+\hat{j}-\hat{k}\) which displace it from a point \(\hat{i}+2\hat{j}+3\hat{k}\) to the point \(5\hat{i}+4\hat{j}+\hat{k}\). The work done in standard units by the forces is given by

Let $\vec{a} = \hat{i}+2\hat{j}+3\hat{k}$, $\vec{b} = \hat{i}-\hat{j}+2\hat{k}$ and $\vec{c} = 5\hat{i}-3\hat{j}+3\hat{k}$ be three vectors. If $\vec{r}$ is a vector such that $\vec{r}\times\vec{b} = \vec{c}\times\vec{b}$ and $\vec{r}\cdot\vec{a}=0$. Then $25|\vec{r}|^2$ is equal to

Let $\vec{a} = -\hat{i}-\hat{j}+\hat{k}$, $\vec{a}\cdot\vec{b} = 1$ and $\vec{a}\times\vec{b} = \hat{i}-\hat{j}$. Then $\vec{a}-6\vec{b}$ is equal to

Let $\vec{a} = 5\hat{i}-\hat{j}-3\hat{k}$ and $\vec{b} = \hat{i}+3\hat{j}+5\hat{k}$ be two vectors. Then which one of the following statements is TRUE?

Unit vector perpendicular to the plane of \(\triangle ABC\) with position vectors \(\vec{a}, \vec{b}, \vec{c}\) of the vertices \(A, B, C\) is

$a\hat{i}+\hat{j}+\hat{k}$, $\hat{i}+b\hat{j}+\hat{k}$, $\hat{i}+\hat{j}+c\hat{k}$ coplanar ($a,b,c\neq1$). Then $\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}$ is equal to

Points $\alpha\hat{i}+10\hat{j}+13\hat{k}$, $6\hat{i}+11\hat{j}+11\hat{k}$, $9\hat{i}+\beta\hat{j}-8\hat{k}$ are collinear. Then $\dfrac{(19\alpha-6\beta)^2}{2}$ is equal to

Let u, v and w be such that \(|\mathbf{u}| = 1\), \(|\mathbf{v}| = 2\), \(|\mathbf{w}| = 3\). If the projection of v along u is equal to that of w along u and v, w are perpendicular to each other, then \(|\mathbf{u} - \mathbf{v} + \mathbf{w}|\) equals

Let \(\alpha \in R\) and the three vectors \(\vec{a} = \alpha\hat{i} + \hat{j} + 3\hat{k}\), \(\vec{b} = 2\hat{i} + \hat{j} - \alpha\hat{k}\) and \(\vec{c} = \alpha\hat{i} - 2\hat{j} + 3\hat{k}\). Then the set \(S = \{\alpha : \vec{a}, \vec{b}\text{ and }\vec{c}\text{ are coplanar}\}\)

Let $\vec{a} = 4\hat{i}+3\hat{j}$ and $\vec{b} = 3\hat{i}-4\hat{j}+5\hat{k}$ and $\vec{c}$ is a vector such that $\vec{c}\cdot(\vec{a}\times\vec{b})+25=0$, $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4$ and projection of $\vec{c}$ on $\vec{a}$ is 1, then the projection of $\vec{c}$ on $\vec{b}$ equals:

If u, v and w are non-coplanar vectors and p, q are real numbers, then the equality [3u pv pw] − [pv w qu] − [2w qv qu] = 0 holds for

The vector $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3\vec{a} + \sqrt{2}\vec{b}$ on $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ is

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that \[\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS} = \overrightarrow{OQ} \cdot \overrightarrow{OR} + \overrightarrow{OP} \cdot \overrightarrow{OS}\]Then the triangle PQR has S as its

[JEE Main 2019] Let \(\vec{a}\) and \(\vec{b}\) be unit vectors and \(\alpha\) be the angle between them. Then \(\vec{a}+\vec{b}\) is a unit vector if

Let a, b and c be three unit vectors, out of which vectors b and c are non-parallel. If α and β are the angles which vector a makes with vectors b and c respectively and a × (b × c) = \frac{1}{2}b, then |a − b| is equal to

Let $\vec{a} = 2\hat{i}-7\hat{j}+5\hat{k}$, $\vec{b} = \hat{i}+\hat{k}$ and $\vec{c} = \hat{i}+2\hat{j}-3\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r}\times\vec{a} = \vec{c}\times\vec{a}$ and $\vec{r}\cdot\vec{b}=0$, then $|\vec{r}|$ is equal to:

Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that $\vec{a}\times\vec{b}=2(\vec{a}\times\vec{c})$. If $|\vec{a}|=1$, $|\vec{b}|=4$, $|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is $60^\circ$, then $|\vec{a}\cdot\vec{c}|$ is equal to:

Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non zero vectors such that $\vec{b}\cdot\vec{c}=0$ and $\vec{a}\times(\vec{b}\times\vec{c}) = \frac{\vec{b}-\vec{c}}{2}$. If $\vec{d}$ be a vector such that $\vec{b}\cdot\vec{d} = \vec{a}\cdot\vec{b}$, then $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})$ is equal to

Let b and c be vectors such that |b × c| = 2 and |b| = |c| = 1. If 2b − c = λa, find α + β where λ = √(α − β√3).

Let a b = 3 i + j - k and c ^ ^ ^ \to and \tob . If the vector C\to \to be three vectors such that c\to is coplanar with a is perpendicular to \tob and a \to ⋅ c\to = 5, then |c\to| is equal to

Let a \to = ^i + ^j + k, ^ \to \times b . If \to b = 2 i + 2 j + k and d = a ^ ^ ^ \to⋅\to c is a vector such that a \to \to \to| = 8 c = | c | , | c - 2a 2 \to \to \to \to \to \to 2 and the angle between d and c is \pi 4 , then |10 - 3 b ⋅ c | + | d \times c | is equal to

Let a and b be two unit vectors. If the vectors c = a + 2b and d = 5a − 4a are perpendicular to each other, then the angle between a and b is

Let $\vec{a}=-\hat{i}+2\hat{j}+2\hat{k}$, $\vec{b}=8\hat{i}+7\hat{j}-3\hat{k}$ and $\vec{c}$ be a vector such that $\vec{a}\times\vec{c}=\vec{b}$ and $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})=4$. Then $|\vec{a}+\vec{c}|^2$ is equal to:

Between the following two statements: Statement I: Let $\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\vec{a}\times\vec{r}=\vec{a}\times\vec{b}$ and $\vec{a}\cdot\vec{r}=0$ is of magnitude $\sqrt{10}$. Statement II: In a triangle $ABC$, $\cos2A+\cos2B+\cos2C\geq-\dfrac{3}{2}$.

We have \(|\vec{a}\times\vec{b} - \vec{a}\times\vec{c}|^2 = |\vec{a}\times(\vec{b}-\vec{c})|^2\). If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors and the angle between \(\vec{b}\) and \(\vec{c}\) is \(\dfrac{\pi}{3}\), and \(\vec{a}\cdot(\vec{b}-\vec{c})=0\), find \(|\vec{a}\times\vec{b}-\vec{a}\times\vec{c}|^2\).

Let c\to be the projection vector of \tob = \lambda i^ + 4k, ^ \to = i^ + 2 j^ + 2k \lambda > 0, on the vector a ^ \to + c\to| = 7, then the area . If |a of the parallelogram formed by the vectors \tob and c\to is ________ \to \to \to

$\vec{u}_1,\vec{u}_2,\vec{u}_3$ coplanar and $\vec{v}_1,\vec{v}_2,\vec{v}_3$ also coplanar (expressions in $a,b,c$). Then $6(a+b+c)$ is equal to

If the components of \to a = \alpha i + \beta j + \gamma k along and perpendicular to b = 3 i + j - k respectively, are ^ ^ ^ ^ ^ ^ 16 11 ^ ^ ^ (3 i + j - k) and 1 11 ^ ^ ^ (-4 i - 5 j - 17k) , then \alpha + \beta + \gamma is equal to : 2 2 2

If the four points, whose position vectors are $3\hat{i}-4\hat{j}+2\hat{k}$, $\hat{i}+2\hat{j}-\hat{k}$, $-2\hat{i}-\hat{j}+3\hat{k}$ and $5\hat{i}-2\alpha\hat{j}+4\hat{k}$ are coplanar, then $\alpha$ is equal to

Given $\vec{b} - 2\vec{c} = \lambda\vec{a}$ where $|\vec{b}| = 2\vec{c}| = |\lambda\vec{a}|$, $|\vec{b}| + 4|\vec{c}| = 4b$, $\vec{c} = \lambda^2\vec{a}$, and $|\vec{b}.\vec{c}| + |\vec{b}.\vec{c}| = |\vec{b}|^2|\vec{c}|^2$. Find the value of $\lambda^2$.

We have \([\vec{a}\times\vec{b}\;\; \vec{b}\times\vec{c}\;\; \vec{c}\times\vec{a}] = \lambda[\vec{a}\vec{b}\cdot\vec{c}]^2\). Find \(\lambda\).

Let the position vectors of three vertices of a triangle be 4p\to + q\to - 3r\to, -5p\to + q\to + 2r\to and 2 p - q + 2 r . If the \to \to \to position vectors of the orthocenter and the circumcenter of the triangle are and \alphap\to + \betaq\to + \gammar\to p +q +r 4 respectively, then \alpha + 2\beta + 5\gamma is equal to :

Let A, B, C be three points in xy-plane, whose position vector are given by \sqrt3^i + ^j, ^i + \sqrt3^j and a^i + (1 - a)^j respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between - -\to - -\to the vectors OA and OB is , then the sum of all the possible values of a is : 9 \sqrt2

A weight of 13 kg is supported by two strings of lengths 5 and 12. Given that \(13^2 = 5^2 + 12^2\), the angle \(\angle AOB = \dfrac{\pi}{2}\). Using equilibrium conditions, \(T_1\) and \(T_2\) are:

Let $\vec{a}\times\vec{c}=\vec{a}\times\vec{b}$. If $\vec{a}\cdot\vec{c}=-12$, $\vec{c}\cdot(\hat{i}-2\hat{j}+\hat{k})=5$, then $\vec{c}\cdot(\hat{i}+\hat{j}+\hat{k})$ is equal to _______

Let $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\lambda\hat{j}+2\hat{k}$, $\lambda\in\mathbb{Z}$. Let $\vec{c}=\vec{a}\times\vec{b}$ and $\vec{d}$ be a vector of magnitude 2 in $yz$-plane. If $|\vec{c}|=\sqrt{53}$, then the maximum possible value of $(\vec{c}\cdot\vec{d})^2$ is equal to:

$\vec{a}$ parallel to intersection of planes through $\hat{i}+\hat{j},\hat{i}+\hat{k}$ and $\hat{i}-\hat{j},\hat{j}-\hat{k}$. $\theta$ angle between $\vec{a}$ and $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$, $\vec{a}\cdot\vec{b}=6$. $(\theta,|\vec{a}\times\vec{b}|)=$

$\vec{a}=3\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=2\hat{i}-3\hat{j}+3\hat{k}$. $\vec{a}=\vec{b}\times\vec{c}$, $|\vec{b}|^2=50$. Then $|72-|\vec{b}+\vec{c}|^2|$ is equal to __________.