If $\theta$ is the angle between $\vec{f}=\hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{g}=2 \hat{i}-3 \hat{j}+a \hat{k}$ and $\sin \theta=\sqrt{\frac{24}{28}}$,then $7 a^2+24 a=$

  • A
    $10$
  • B
    $12$
  • C
    $36$
  • D
    $15$

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Similar Questions

Let $\overrightarrow{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$.
Assertion $(A)$ : The identity $|\overrightarrow{a} \times \hat{i}|^2+|\overrightarrow{a} \times \hat{j}|^2+|\overrightarrow{a} \times \hat{k}|^2=2|\overrightarrow{a}|^2$ holds for $\overrightarrow{a}$.
Reason $(R)$ : $\overrightarrow{a} \times \hat{i}=a_3 \hat{j}-a_2 \hat{k}$,$\overrightarrow{a} \times \hat{j}=a_1 \hat{k}-a_3 \hat{i}$,and $\overrightarrow{a} \times \hat{k}=a_2 \hat{i}-a_1 \hat{j}$.
Which of the following is correct?

The area of the parallelogram whose adjacent sides are $\vec{a} = \hat{i} - \hat{k}$ and $\vec{b} = 2\hat{j} + 3\hat{k}$ is

The area of the triangle whose vertices are $(1, 2, 3)$,$(2, 5, -1)$,and $(-1, 1, 2)$ is:

If the position vectors of the vertices of a $\triangle ABC$ are $\vec{OA} = 3\hat{i} + \hat{j} + 2\hat{k}$,$\vec{OB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OC} = 2\hat{i} + 3\hat{j} + \hat{k}$,then the length of the altitude of $\triangle ABC$ drawn from $A$ is

Let $\vec{a}=2 \hat{i}-3 \hat{j}-5 \hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}-5 \hat{k}$ be two vectors and $\vec{r}$ be a vector in the plane of $\vec{a}$ and $\vec{b}$. If $\vec{r}$ is orthogonal to the vector $5 \hat{i}-2 \hat{j}+3 \hat{k}$ and the magnitude of $\vec{r}$ is $\sqrt{94}$,then $|\vec{r} \cdot \vec{b}|=$

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