(D) Given $\bar{a} = (x, -3, -1)$ and $\bar{b} = (2x, x, -1)$.$1$. The angle between $\bar{a}$ and $\bar{b}$ is acute,so $\bar{a} \cdot \bar{b} > 0$.$

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(D) Given $\bar{a} = (x, -3, -1)$ and $\bar{b} = (2x, x, -1)$.$1$. The angle between $\bar{a}$ and $\bar{b}$ is acute,so $\bar{a} \cdot \bar{b} > 0$.$(x)(2x) + (-3)(x) + (-1)(-1) > 0$$2x^2 - 3x + 1 > 0$$(2x - 1)(x - 1) > 0$This implies $x 1$. ... $(1)$$2$. The angle between $\bar{b}$ and the $y$-axis (vector $\bar{j} = (0, 1, 0)$) is obtuse,so $\bar{b} \cdot \bar{j} $(2x, x, -1) \cdot (0, 1, 0) $0 + x + 0 $x From $(1)$ and $(2)$,the intersection is $x < 0$.

Class 12 Mathematics Vector Algebra Scalar or Dot product of two vectors and its applications

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If the angle between $\bar{a} = x\bar{i} - 3\bar{j} - \bar{k}$ and $\bar{b} = 2x\bar{i} + x\bar{j} - \bar{k}$ is acute and the angle between $\bar{b}$ and the $y$-axis is obtuse,then $x$ belongs to which interval?

A
$x < 1/2$
B
$x > 1$
C
$x > 1/2$
D
$x < 0$

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Vector Algebra

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Scalar or Dot product of two vectors and its applications

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then $\cot^2 \theta=$

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The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

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The values of $a$,for which points $A, B, C$ with position vectors $2\hat{i}-\hat{j}+\hat{k}$,$\hat{i}-3\hat{j}-5\hat{k}$,and $a\hat{i}-3\hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $m\angle C = 90^\circ$ are:

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Statement $(A):$ If $|\vec{a}| = 2, |\vec{b}| = 3, |2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}| = 5$.
Reason $(R): |\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$

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If $a \cdot i = a \cdot (i + j) = a \cdot (i + j + k)$, then $a = $

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If the angle between $\bar{a} = x\bar{i} - 3\bar{j} - \bar{k}$ and $\bar{b} = 2x\bar{i} + x\bar{j} - \bar{k}$ is acute and the angle between $\bar{b}$ and the $y$-axis is obtuse,then $x$ belongs to which interval?

Similar Questions

If $|\vec{a}|=4, |\vec{b}|=5, |\vec{a}-\vec{b}|=3$ and $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$, then $\cot^2 \theta=$

The value of $b$ such that the scalar product of the vector $(i + j + k)$ with the unit vector parallel to the sum of the vectors $(2i + 4j - 5k)$ and $(bi + 2j + 3k)$ is $1$,is:

The values of $a$,for which points $A, B, C$ with position vectors $2\hat{i}-\hat{j}+\hat{k}$,$\hat{i}-3\hat{j}-5\hat{k}$,and $a\hat{i}-3\hat{j}+\hat{k}$ respectively are the vertices of a right-angled triangle with $m\angle C = 90^\circ$ are:

Statement $(A):$ If $|\vec{a}| = 2, |\vec{b}| = 3, |2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}| = 5$.Reason $(R): |\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$

If $a \cdot i = a \cdot (i + j) = a \cdot (i + j + k)$, then $a = $

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Statement $(A):$ If $|\vec{a}| = 2, |\vec{b}| = 3, |2\vec{a} - \vec{b}| = 5$,then $|2\vec{a} + \vec{b}| = 5$.
Reason $(R): |\vec{p} - \vec{q}| = |\vec{p} + \vec{q}|$