Vector Algebra Questions - JEE

Consider the regular hexagon ABCDEF with centre at O (origin). Find $\vec{AD} + \vec{EB} + \vec{FC}$

From the parallelogram law of forces, if $R^2 = P^2 + Q^2 + 2PQ\cos\alpha$ and equations are set up for doubled forces, the ratio $P^2 : Q^2 : R^2$ equals:

If ABCD is a parallelogram and the position vectors of A, B and C are $\vec{i} + 3\vec{j} + 5\vec{k}$, $\vec{i} + \vec{j} + \vec{k}$ and $7\vec{i} + 7\vec{j} + 7\vec{k}$, then the position vector of D will be

The horizontal force and the force inclined at an angle 60° with the vertical, whose resultant is in vertical direction of P kg, are

If the vectors $6\mathbf{i}, 2\mathbf{j}$ and $3\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}$ form a triangle, determine the type of triangle.

Let a = 3i − 6j + k and b = 2i − 4j + lk. Since a and b are parallel, find the value of l.

If position vector of point A is a + 2b and a divides AB in the ratio 2 : 3, then the position vector of B is

If a and b are the position vectors of A and B respectively, then the position vector of a point C on AB produced such that AC = 3 AB is

We have a, b, c are non-zero vectors and $\vec{a} \neq \lambda_1 \vec{b}$ and $\vec{b} \neq \mu_1 \vec{c}$, $\lambda_1, \mu_1 \neq 0$. Given:$\vec{a} + 2\vec{b} = m\vec{c}$ ...(1)and $\vec{b} + 3\vec{c} = n\vec{a}$ ...(2)where $m, n \neq 0$ are reals. Find $\vec{a} + 2\vec{b} + 6\vec{c}$ = ?

If $a$, $b$ and $c$ are $p$th, $q$th, $r$th terms of HP and $\vec{u} = (q - r)\vec{i} + (r - p)\vec{j} + (p - q)\vec{k}$, $\vec{v} = \frac{1}{a}\vec{i} + \frac{1}{b}\vec{j} + \frac{1}{c}\vec{k}$, then

The angles of a triangle, two of whose sides are represented by vectors 3(a × b) and b - (a · b)a, where b is a non-zero vector and a is a unit vector in the direction of a, are

Direction of the ant's resultant displacement after first three steps is :

If the position vectors of the points $A$, $B$ and $C$ be $\vec{i} + \vec{j}$, $\vec{i} - \vec{j}$ and $a\vec{i} + b\vec{j} + c\vec{k}$ respectively, then the points $A$, $B$ and $C$ are collinear, if

A vector $\vec{C}$ directed along internal bisector of angle between vectors $\vec{A} = 7\vec{i} - 4\vec{j} - 4\vec{k}$ and $\vec{B} = -2\vec{i} - \vec{j} + 2\vec{k}$ with $|\vec{C}| = 5\sqrt{6}$ is

The sine of the angle between the vectors a = 3i + j + k and b = 2i - 2j + k is

In a parallelogram ABCD, $\vec{AB} = \vec{i} + \vec{j} + \vec{k}$ and diagonal $\vec{AC} = \vec{i} - \vec{j} + \vec{k}$ and area of parallelogram is 8 sq units, then $\angle BAC$ is equal to

Let $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a}\times\vec{b}$. If $\vec{v}\cdot\vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9p^2$ is equal to:

Let $\vec{u} = \hat{i} - \hat{j} - 2\hat{k}$, $\vec{v} = 2\hat{i} + \hat{j} - \hat{k}$, $\vec{v}\cdot\vec{w} = 2$ and $\vec{v}\times\vec{w} = \vec{u} + \lambda\vec{v}$. Then $\vec{u}\cdot\vec{w}$ is equal to

Let $\vec{\alpha} = 4\hat{i} + 3\hat{j} + 5\hat{k}$ and $\vec{\beta} = \hat{i} + 2\hat{j} - 4\hat{k}$. Let $\vec{\beta}_1$ be parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ be perpendicular to $\vec{\alpha}$. If $\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2$, then the value of $5\vec{\beta}_2\cdot(\hat{i}+\hat{j}+\hat{k})$ is

Let $\vec{a} = \hat{i} + 2\hat{j} + \lambda\hat{k}$, $\vec{b} = 3\hat{i} - 5\hat{j} - \lambda\hat{k}$, $\vec{a}\cdot\vec{c} = 7$, $2\vec{b}\cdot\vec{c} + 43 = 0$, $\vec{a}\times\vec{c} = \vec{b}\times\vec{c}$. Then $|\vec{a}\cdot\vec{b}|$ is equal to

Let $\lambda \in \mathbb{R}$, $\vec{a} = \lambda\hat{i}+2\hat{j}-3\hat{k}$, $\vec{b} = \hat{i}-\lambda\hat{j}+2\hat{k}$. If $((\vec{a}+\vec{b})\times(\vec{a}\times\vec{b}))\times(\vec{a}-\vec{b}) = 8\hat{i}-40\hat{j}-24\hat{k}$, then $|\lambda(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})|^2$ is equal to

Magnitude of the resultant displacement is given by :

Two adjacent sides of a parallelogram ABCD are given by $\overrightarrow{AB} = 2\vec{i} + 10\vec{j} + 11\vec{k}$ and $\overrightarrow{AD} = -\vec{i} + 2\vec{j} + 2\vec{k}$. The side AD is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that AD becomes $\overrightarrow{AD'}$. If $\overrightarrow{AD'}$ makes a right angle with the side AB, then the cosine of the angle $\alpha$ is given by

Ex. 46 Statement I: If $|\mathbf{a} + \mathbf{b}| = |\mathbf{a} - \mathbf{b}|$, then $\mathbf{a}$ and $\mathbf{b}$ are perpendicular to each other.Statement II: If the diagonals of a parallelogram are equal in magnitude, then the parallelogram is a rectangle.

Let a = $2\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ and b = $\mathbf{i} + \mathbf{j}$. If c is a vector such that $\mathbf{a} \cdot \mathbf{c} = |\mathbf{c}|$, $|\mathbf{c} - \mathbf{a}| = 2\sqrt{2}$ and the angle between $\mathbf{a} \times \mathbf{b}$ and c is 30°, then $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|$ is equal to

The non-zero vectors a, b and c are related by a = 8b and c = -7b. Find the angle between a and c.

If vectors $\vec{a} = \frac{\vec{i} + \vec{j}}{\sqrt{2}}, \vec{b} = \frac{-\vec{i} + \vec{j}}{\sqrt{2}}$ and $\vec{c} = \vec{k}$ then the value of $(\vec{r} \cdot \vec{a})^2 + (\vec{r} \cdot \vec{b})^2 + (\vec{r} \cdot \vec{c})^2$ is equal to:

Ex. 56 In the regular pentagon ABCDE with the given position vectors, AD divides EC in the ratio

If the volume of parallelepiped formed by the vectors a, b, c as three coterminous edges is 27 cu units, then the volume of the parallelepiped with $\alpha = a + 2b - c$, $\beta = a - b$ and $\gamma = a - b - c$ as three coterminous edges is

The position vectors of the points A, B and C are i + 2j - k, i + j + k, and i + 3j + 2k, respectively. If A is [incomplete question]

If O be the circumcentre and O' be the orthocentre of the $\triangle ABC$, then $\vec{O'A} + \vec{O'B} + \vec{O'C}$ is equal to

A line segment has length 63 and direction ratios are 3, -2 and 6. Find the components of the line vector.

The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is

$(\vec{a} \times \vec{b}) \times \vec{c} = \vec{a} \times (\vec{b} \times \vec{c})$ is $\vec{a}$ and $\vec{c}$ are:

Let the cosine of angle between the vectors p and q be $λ$ such that $2\mathbf{p} + \mathbf{q} = \mathbf{i} + \mathbf{j}$ and $\mathbf{p} + 2\mathbf{q} = \mathbf{i} - \mathbf{j}$, then $λ$ is equal to

The position vector of the points which divides internally in the ratio 2 : 3 the join of the points 2a - 3b and 3a - 2b, is

Consider points $A$, $B$, $C$ and $D$ with position vectors $7\hat{i} - 4\hat{j} + 7\hat{k}$, $\hat{i} - 6\hat{j} + 10\hat{k}$, $-\hat{i} - 3\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + \hat{k}$, respectively. Then $ABCD$ is a

Let $a$, $b$, $c$ be three vectors such that $[a b c] = 2$. If $r = l(b \times c) + m(c \times a) + n(a \times b)$ is perpendicular to $a + b + c$, then the value of $(l + m + n)$ is

Let $\vec{a} = \vec{i} + 2\vec{j} + \vec{k}$, $\vec{b} = \vec{i} - \vec{j} + \vec{k}$, $\vec{c} = \vec{i} + \vec{j} - \vec{k}$. A vector coplanar to $\vec{a}$ and $\vec{b}$ has a projection along $\vec{c}$ of magnitude $\frac{1}{3}$. Then the vector is:

Given that $\vec{a}$ is perpendicular to $\vec{b}$ and p is a non-zero scalar, if $p\vec{r} + (\vec{r} \cdot \vec{b})\vec{a} = \vec{c}$ then $\vec{r} = $:

If $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are on a circle of radius $R$ whose centre is at origin and $\vec{c} - \vec{a}$ is perpendicular to $\vec{d} - \vec{b}$, then $|\vec{d} - \vec{a}|^2 + |\vec{b} - \vec{c}|^2$ (AC is diameter)

Let two non-collinear unit vectors $\vec{a}$ and $\vec{b}$ form an acute angle. A point P moves, so that at any time $t$ the position vector $\overrightarrow{OP}$ (where O is the origin) is given by $\overrightarrow{OP} = \vec{a}\cos t + \vec{b}\sin t$. When P is farthest from origin O, let M be the length of $\overrightarrow{OP}$ and $\vec{u}$ be the unit vector along $\overrightarrow{OP}$. Then find $\vec{u}$ and $M$.

If the position vectors of the points A and B are i + 3j - k and 3i - j - 3k, then what will be the position vector of the mid-point of AB?

The vector $\vec{OP} = 5\vec{i} + 12\vec{j} + 13\vec{k}$ turns through an angle of $\frac{\pi}{2}$ about $O$ passing through the positive side of $\vec{j}$ axis an iff way. The vector in the new position is:

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three non-parallel unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2}\vec{b}$, the angle between $\vec{b}$ and $\vec{c}$ is:

Example 32. The length of longer diagonal of the parallelogram constructed on $5\mathbf{a} + 2\mathbf{b}$ and $\mathbf{a} - 3\mathbf{b}$, when it is given that $|\mathbf{a}| = 2\sqrt{2}$, $|\mathbf{b}| = 3$ and angle between $\mathbf{a}$ and $\mathbf{b}$ is $\frac{\pi}{4}$, is

Let r, a, b and c be four non-zero vectors such that r · a = 0, |r × b| = |r| |b|, |r × c| = |r| |c|, then [a b c] is

Given $\vec{\alpha} = (\lambda - 2)\vec{a} + \vec{b}$ and $\vec{\beta} = (4\lambda - 2)\vec{a} + 3\vec{b}$. Vectors $\vec{\alpha}$ and $\vec{\beta}$ are collinear. Find $\lambda$.

What should be added to vector a = 3i + 4j - 2k to get its resultant a unit vector i?

Let $a, b, c$ denote the lengths of the sides of a triangle such that $(a-b)\vec{u} + (b-c)\vec{v} + (c-a)(\vec{u} \times \vec{v}) = 0$ for any two non-collinear vectors $\vec{u}$ and $\vec{v}$, then the triangle is

Vector Algebra Questions (573)