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:
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:
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
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.
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:
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.
\(A,B,C,D\) are four points in a plane with position vectors \(\vec{a},\vec{b},\vec{c},\vec{d}\) respectively such that \((\vec{a}-\vec{d})\cdot(\vec{b}-\vec{c})=0\) and \((\vec{b}-\vec{d})\cdot(\vec{c}-\vec{a})=0\). Then \(D\) is the
For \(p>0\), the vector \(\vec{v}_2=(2,\,-(\sqrt{3}\,p+1))\) is obtained by rotating \(\vec{v}_1=\sqrt{3}(-1,\,-(p^2+\sqrt{3}))\) about the origin counter-clockwise. If the angle of rotation is \(\theta\), find \(\tan\theta\).
Let a vector \(\hat{i}+\sqrt{2}\,\hat{j}+\sqrt{2}\,\hat{k}\) be obtained by rotating the vector \(\sqrt{3}\,\hat{j}\) by an angle \(45°\) about the origin in the clockwise direction to the first quadrant. Then the area of the triangle formed by the vector \((\hat{i}+\sqrt{2}\,\hat{j}+\sqrt{2}\,\hat{k})\) with the coordinate axes is equal to:
Consider points \(A,B,C\) with position vectors \(\vec{a},\vec{b},\vec{c}\) respectively. Statement-1: \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}\). Statement-2: \(A,B,C\) form the vertices of a triangle.