MHT CET 2025 22nd April Evening Shift | Vector Algebra Question 4 | Physics

Given $\quad \vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k}), \quad \vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})$ and $\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k}) \cdot(\vec{A}+\vec{B}) \cdot \vec{C}$ will be

12 c

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

Given $\quad \vec{A}=(2 \hat{i}-3 \hat{j}+\hat{k}), \quad \vec{B}=(3 \hat{i}+\hat{j}-2 \hat{k})$ and $\vec{C}=(3 \hat{i}+2 \hat{j}+\hat{k}) \cdot(\vec{A}+\vec{B}) \cdot \vec{C}$ will be

12 c

MHT CET 2025 21st April Evening Shift

MCQ (Single Correct Answer)

-0

A unit vector in the direction of resultant vector of $\vec{A}=-2 \hat{i}+3 \hat{j}+\hat{k}$ and $\vec{B}=\hat{i}+2 \hat{j}-4 \hat{k}$ is

$\frac{-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{35}}$

$\frac{\hat{i}+2 \hat{j}-4 \hat{k}}{\sqrt{35}}$

$\frac{-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{35}}$

$\frac{-\hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}}{\sqrt{35}}$

MHT CET 2025 21st April Morning Shift

MCQ (Single Correct Answer)

-0

The three vector $\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k}, \vec{B}=\hat{i}-3 \hat{j}+5 k$ and $\vec{C}=2 \hat{i}-\hat{j}+4 \hat{k}$ will form

isosceles triangle.

equilateral triangle.

no triangle.

right angled triangle.

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