Vector Algebra Questions (573)

The foot of perpendicular from the origin O to a plane P which meets the co-ordinate axes at the points A, B, C is $(2, a, 4)$, $a \in \mathbb{N}$. If the volume of the tetrahedron OABC is 144 unit$^3$, then which of the following points is NOT on P?

If three coterminous edges of a tetrahedron are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 4$, angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$, $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{4}$ and $\vec{c}$ and $\vec{a}$ is $\frac{\pi}{6}$. The area of the base is $2$ sq. units, then the height of the tetrahedron is:

If $\vec{\alpha}$ and $\vec{\beta}$ be two perpendicular unit vectors such that $\vec{x} = \vec{\beta} - (\vec{a} \times \vec{x})$, then $|\vec{x}|$ is equal to:

Let \(\vec{a}=\hat{i}+\hat{j}+2\hat{k}\), \(\vec{b}=2\hat{i}-\hat{j}+\hat{k}\), \(\vec{c}=3\hat{i}-\hat{k}\). A vector \(\vec{v}\) is coplanar with \(\vec{a}\) and \(\vec{b}\), perpendicular to \(\vec{c}\), and satisfies \(\vec{v}\cdot(\hat{i}+2\hat{j}+\hat{k})=8\). Find \(|\vec{v}|^2\).

Let $\vec{a} = a\vec{i} + b\vec{j} + c\vec{k}$ and $\vec{\beta} = b\vec{i} + \vec{c}\vec{j} + a\vec{k}$, where $a, b, c \in \mathbb{R}$. If '$\theta$' be the angle between $\vec{a}$ and $\vec{\beta}$ then:

Let \(OAB\) be a regular triangle (equilateral) with \(O\) at the origin. If \(\overrightarrow{OA}=\vec{a}\) and \(\overrightarrow{OB}=\vec{b}\), then which of the following hold?

A vector $\vec{a} = a\vec{i} + b\vec{j} + c\vec{k}$ is said to be rational vector. If $a, b, c$ are all rational. If a rational vector with magnitude as positive integer makes an angle $\pi/4$ with vector $\vec{b} = \sqrt{2}\vec{i} + \sqrt{2}\vec{j} + \vec{k}$, then $\vec{a}$:

$ABC$ is a triangle. $AD, AD'$ are internal and external bisectors of angle $A$, meeting $BC$ at $D$ and $D'$ respectively. $A'$ is the mid-point of $DD'$ and $B', C'$ are similar points on $CA$ and $AB$. Then $A', B', C'$:

Let $\vec{a} = \vec{i} + \vec{j}$ and $\vec{b} = 2\vec{i} - \vec{k}$. The point of intersection of the lines $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is:

If for a plane, $x$, $y$, $z$ intercepts are $8$, $4$ respectively, then the length of the perpendicular from the origin on to the plane is

If $\vec{a},\vec{b},\vec{c},\vec{d}$ are coplanar, then $[\vec{a}\vec{b}\vec{c}]$ is equal to

If $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) \cdot (\vec{c} \times \vec{b}) = 0$ then which of the following is always true:

For three unit vectors $\vec{a},\vec{b},\vec{c}$ satisfying $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2=9$ and $|2\vec{a}+k\vec{b}+k\vec{c}|=3$, the positive value of $k$ is:

Let $\vec{u}$, $\vec{v}$ and $\vec{w}$ are vectors such that $\vec{u} + \vec{v} + \vec{w} = \vec{0}$. If $|\vec{u}| = 3$, $|\vec{v}| = 4$ and $|\vec{w}| = 5$, then the value of $\vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{u}$ is

The position vectors of the vertices $A, B$ and $C$ of a triangle are three unit vectors $\vec{a}, \vec{b}$ and $\vec{c}$. A vector $\vec{d}$ is such that $\vec{d} \cdot \vec{a} = \vec{d} \cdot \vec{b} = \vec{d} \cdot \vec{c}$ and $\vec{d} = \lambda(\vec{b} + \vec{c})$, then triangle $ABC$ is:

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-coplanar vectors and $\vec{d}$ be a non-zero vector, which is perpendicular to $(\vec{a} + \vec{b} + \vec{c})$. Now if $\vec{d} = \sin x(\vec{a} \times \vec{b}) + \cos y(\vec{b} \times \vec{c}) + 2(\vec{c} \times \vec{a})$, then minimum value of $x^2 + y^2$ is equal to:

Let position vector of point $A$ be $\vec{i} + \vec{j} + \vec{k}$ and that of point $B$ be $-\vec{i} + \vec{k}$, then the position vector of point $R(\vec{r})$ such that $AR$ is perpendicular to $BR$ and $\vec{r}$ is not perpendicular to $\vec{r} - (\vec{j} + 2\vec{k})$ is:

A vector \(\vec{a}\) has components \(2p\) and \(1\) w.r.t. a rectangular Cartesian system. This system is rotated through a certain angle about the origin (counter-clockwise). In the new system the components are \(p+1\) and \(\sqrt{3}\). Then a value of \(p\) is equal to:

If the vectors \(\vec{a} = \hat{i} - \hat{j} + 2\hat{k}\), \(\vec{b} = 2\hat{i} + 4\hat{j} + \hat{k}\) and \(\vec{c} = \lambda\hat{i} + \hat{j} + \mu\hat{k}\) are mutually orthogonal, then \((\lambda, \mu) =\)

Let $ABC$ be a triangle and $\vec{a},\vec{b},\vec{c}$ be the position vectors of the point $A, B, C$ respectively. External bisectors of $\angle B$ and $\angle C$ meet at $P$ with the sides of the triangle as $a, b, c$, the position vector of $P$ becomes :

If vectors $\vec{b} = (\tan\alpha, -1, 2\sqrt{\sin\frac{\alpha}{2}})$ and $\vec{c} = (\tan\alpha, \tan\alpha, -\frac{3}{\sqrt{\sin\alpha/2}})$ are orthogonal and vector $\vec{a} = (1, 3, \sin2\alpha)$ makes an obtuse angle with the z-axis then:

If $\vec{r} = l(\vec{b} \times \vec{c}) + m(\vec{c} \times \vec{a}) + n(\vec{a} \times \vec{b})$ and $[\vec{b}\vec{c}\vec{c}] = 2$, then $l + m + n$ is equal to:

A vector $\vec{c}$, directed along the internal bisector of the angle between the 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 :

Let $\vec{a}=2\hat{i}-\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+3\hat{j}-\hat{k}$ and $\vec{c}=2\hat{i}+\hat{j}+3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\dfrac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to:

Let $\overrightarrow{AB}=2\hat{i}+4\hat{j}-5\hat{k}$ and $\overrightarrow{AD}=\hat{i}+2\hat{j}+\lambda\hat{k}$, $\lambda\in\mathbb{R}$. Let the projection of the vector $\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal $\overrightarrow{AC}$ of the parallelogram ABCD be of length one unit. If $\alpha,\beta$, where $\alpha>\beta$, be the roots of the equation $\lambda^2x^2-6\lambda x+5=0$, then $2\alpha-\beta$ is equal to:

Three forces $\vec{F}_1 = \vec{i} + 2\vec{j} - 3\vec{k}$, $\vec{F}_2 = 2\vec{i} + 3\vec{j} + 4\vec{k}$ and $\vec{F}_3 = \vec{i} - \vec{j} + \vec{k}$ acting on a particle at the point $(0, 1, 2)$. The magnitude of the moment of the forces about the point $(1, -2, 0)$ is

If \(|\vec{a}| = 3\), \(|\vec{b}| = 4\), then find a value of \(\lambda\) for which \(\vec{a} + \lambda \vec{b}\) is perpendicular to \(\vec{a} - \lambda \vec{b}\).

Given two vectors \(\hat{i} - \hat{j}\) and \(\hat{i} + 2\hat{j}\), the unit vector coplanar with the two vectors and perpendicular to the first is

The direction cosines of the vector $3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ are

a = \(2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), b = \(-\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\), c = \(\mathbf{i} + \mathbf{j} + \mathbf{k}\) and d = \(3\mathbf{i} + 2\mathbf{j} + \mathbf{k}\), then \(\frac{1}{7}(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d})\) is equal to:

The points \(\mathbf{i} - \mathbf{j} + 3\mathbf{k}\) and \(3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}\) are equidistant from the plane \(\mathbf{r} \cdot (5\mathbf{i} + 2\mathbf{j} - 7\mathbf{k}) + 9 = 0\), then they are

Given |a| = |b| = |c| = 1 and |a − b|² + |b − c|² + |c − a|² = 9, find |2a + 5b + 5c|.

Example 31. The vectors c, a = xi + yj + zk and b = j are such that a, c, and b form a right handed system. Then c is:

If u, v and w are three non-coplanar vectors, then \((\mathbf{u} + \mathbf{v} - \mathbf{w}) \cdot [(\mathbf{u} - \mathbf{v}) \times (\mathbf{v} - \mathbf{w})]\) is equal to

If a = i + j + k, b = i - j + k, c = i + 2j - k, then the value of \[\begin{vmatrix} \mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c} \end{vmatrix}\] is

Volume of parallelopiped with edges a, b and c is

Let \mathbf{a}, \mathbf{b} > 0 and \boldsymbol{\alpha} = \frac{4}{a}\mathbf{i} + \frac{j}{b} + \mathbf{b}\mathbf{k} and \boldsymbol{\beta} = \mathbf{b}\mathbf{i} + \mathbf{a}\mathbf{j} + \frac{1}{b}\mathbf{k}, then the maximum value of \frac{10}{5 + \boldsymbol{\alpha} \cdot \boldsymbol{\beta}} is

Let \(a\), \(b\) and \(c\) be distinct non-negative numbers. If the vectors \(a\hat{i} + a\hat{j} + c\hat{k}\), \(\hat{i} + \hat{k}\) and \(c\hat{i} + c\hat{j} + b\hat{k}\) lie in a plane, then \(c\) is

A particle moves towards east from a point A to a point B at the rate of 4 km/h and then towards north from B to C at the rate of 5 km/h. If \(AB = 12\) km and \(BC = 5\) km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are, respectively,

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 \(|\vec{u}-\vec{v}+\vec{w}|\) equals

As \(\vec{a} = 8\vec{b}\), we have \(\vec{c} = -7\vec{b}\). Therefore, \(\vec{a}\) and \(\vec{b}\) are like vectors and \(\vec{b}\) and \(\vec{c}\) are unlike. This implies that \(\vec{a}\) and \(\vec{c}\) will be unlike. The angle between \(\vec{a}\) and \(\vec{c}\) is equal to:

Given vectors p, q, and r such that p = \frac{2}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}, q = -\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}, and r = -\frac{1}{3}\mathbf{i} - \frac{1}{3}\mathbf{j} + \frac{2}{3}\mathbf{k}. If 3(\mathbf{p} \times \mathbf{q})^2 - \lambda|\mathbf{r} \times \mathbf{q}| = 0, find the value of \lambda.

If |\mathbf{a}| = 5, |\mathbf{a} - \mathbf{b}| = 8 and |\mathbf{a} + \mathbf{b}| = 10, then |\mathbf{b}| is equal to

We have \(\vec{a} \cdot \vec{b} = 0,\ \vec{b} \cdot \vec{c} = 0,\ \vec{c} \cdot \vec{a} = 0\). Given \(2\lambda + 4 + \mu = 0\) and \(\lambda - 1 + 2\mu = 0\). Find the values of \(\lambda\) and \(\mu\).

Since ā, c̄, b̄ form a right handed system, find c̄ given that b̄ = (0, 1, 0) and ā = (x, y, z).If ā, c̄, b̄ form a right handed system and \(\vec{c} = \vec{b} \times \vec{a}\), then \(\vec{c}\) equals:

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors such that no two of these are collinear. If the vector \(\vec{a} + 2\vec{b}\) is collinear with \(\vec{c}\) and \(\vec{b} + 3\vec{c}\) is collinear with \(\vec{a}\) (\(\lambda\) being some non-zero scalar) then \(\vec{a} + 2\vec{b} + 6\vec{c}\) equals

Let \(ABCD\) be a parallelogram such that \(\overrightarrow{AB} = \vec{q}\), \(\overrightarrow{AD} = \vec{p}\) and \(\angle BAD\) be an acute angle. If \(\vec{r}\) is the vector that coincides with the altitude directed from the vertex \(B\) to the side \(AD\), then \(\vec{r}\) is given by

If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are unit vectors such that \(\vec{a}+2\vec{b}+2\vec{c}=\vec{0}\), then \(|\vec{a}\times\vec{c}|\) is equal to

If \(\vec{a}\) and \(\vec{b}\) are non-zero, non-collinear vectors and \(\vec{a_1} = \lambda \vec{a} + 3\vec{b}; \vec{b_1} = 2\vec{a} + \lambda \vec{b}; \vec{c_1} = \vec{a} + \vec{b}\). Find the sum of all possible real values of \(\lambda\) so that points \(A_1, B_1, C_1\) whose position vectors are \(\vec{a_1}, \vec{b_1}, \vec{c_1}\) respectively are collinear.

Two forces \(\vec{f_1} = 3\vec{i} - 2\vec{j} + \vec{k}\) and \(\vec{f_2} = 2\vec{i} + 3\vec{j} - 5\vec{k}\) acting on a particle at A move it to B. The work done, if the position vector of A and B are \(-2\vec{i} + 5\vec{k}\) and \(3\vec{i} - 7\vec{j} + 2\vec{k}\), is