Download App

Scalar Triple Product — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsScalar Triple Product1 Mark
MCQ
If the vectors $4\hat{\text{i}}+11\hat{\text{j}}+\text{m}\hat{\text{k}},7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$ and $\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$ are coplanar, then $m =$
  • A
    $0$
  • B
    $38$
  • C
    $-10$
  • ✓
    $10$
✓

Answer

Correct option: D.
$10$
Let
$\vec{\text{a}}=4\hat{\text{i}}+11\hat{\text{j}}+{\text{m}}\hat{\text{k}}$
$\vec{\text{b}}=7\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\vec{\text{c}}=\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}}$
We know that vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are coplanar if their scalar triple product is zero, i.e. $\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big]=0$
$\Rightarrow\begin{vmatrix}4&11&\text{m}\\7&2&6\\1&5&4 \end{vmatrix}=0$
$\Rightarrow 4(8-30)-11(28-6)+\text{m}(35-2)=0$
$\Rightarrow-88-242+33\text{m}=0$
$\Rightarrow33\text{m}=330$
$\therefore\text{m}=10$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Scalar Triple Product→Scalar Triple Product — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

The inverse of $A=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]$
→
If $\bar{a}, \bar{b}$ are unit vectors such that the vector $\overline{ a }+3 \overline{b}$ is perpendicular to $7 \overline{ a }-5 \overline{b}$ and $\overline{ a }-4 \overline{b}$ is perpendicular to $7 \bar{a}-2 \bar{b}$, then the angle between $\bar{a}$ and $\bar{b}$ is
→
Let $a, b, c$ be positive real numbers. The following system of equations in $x, y$ and $z \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}-\frac{\text{z}^2}{\text{c}^2}=1,$ $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1,$ $-\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{\text{z}^2}{\text{c}^2}=1$ has:
→
The acute angle between the lines represented by $x^2+x y=0$ is
→
The value of integral $\int_{1 / \pi}^{2 / \pi} \frac{\sin \left(\frac{1}{x}\right)}{x^2} d x=$
→

Evaluate: $\int \frac{x^3+x}{x^4-9} d x$

→
The value of the definite integral $\int_0^1 \frac{d x}{x^2+2 x \cos \alpha+1}$, for $0<\alpha<\pi$ is equal to
→
If $\overline{ a }, \overline{ b }, \overline{ c }$ be any three non-coplanar vectors, then $[\bar{a}+\bar{b} \ \bar{b}+\bar{c} \ \bar{c}+\bar{a}]=$
→
The value of the determinant $\begin{vmatrix}\text{a}^2&\text{a}&1\\\cos\text{nx}&\cos(\text{n}+1)\text{x}&\cos(\text{n}+2)\text{x}\\\sin\text{nx}&\sin(\text{n}+1)\text{x}&\sin(\text{n}+2)\text{x}\end{vmatrix}$ is independent of:
→
A solution set of the inequality $x \geq 0$
→