Prashnikaप्रश्निका

Vectors - ISC Class 12 Mathematics Questions with Answers, Page 4

65 past-paper questions on Vectors from ISC Class 12 Mathematics papers (2026-2017), newest first, in full. Questions 61-65 are on this page, 20 to a page. Tap "Show answer" under a question to see its answer.

Practise these questions with filters
2018 · 4 marks · DerivationOpen: Show that the four points and with position vectors , , and respectively, are…
Show that the four points $A, B, C$ and $D$ with position vectors $4\hat{\imath} + 5\hat{\jmath} + \hat{k}$, $-\hat{\jmath} - \hat{k}$, $3\hat{\imath} + 9\hat{\jmath} + 4\hat{k}$ and $4(-\hat{\imath} + \hat{\jmath} + \hat{k})$ respectively, are coplanar.

Given: Four points $A, B, C, D$ with position vectors $4\hat{\imath} + 5\hat{\jmath} + \hat{k}$, $-\hat{\jmath} - \hat{k}$, $3\hat{\imath} + 9\hat{\jmath} + 4\hat{k}$ and $4(-\hat{\imath} + \hat{\jmath} + \hat{k})$

To show: Points $A, B, C, D$ are coplanar

No answer yet.

2018 · 6 marks · Short answerOpen: [3×2] Find if the scalar projection of on is units. The Cartesian equation of a…
[3×2]
(a)[2.0]
Find $\lambda$ if the scalar projection of $\vec{a} = \lambda\hat{\imath} + \hat{\jmath} + 4\hat{k}$ on $\vec{b} = 2\hat{\imath} + 6\hat{\jmath} + 3\hat{k}$ is $4$ units.
(b)[2.0]
The Cartesian equation of a line is: $2x - 3 = 3y + 1 = 5 - 6z$. Find the vector equation of a line passing through $(7, -5, 0)$ and parallel to the given line.
(c)[2.0]
Find the equation of the plane through the intersection of the planes $\vec{r} \cdot (\hat{\imath} + 3\hat{\jmath} - \hat{k}) = 9$ and $\vec{r} \cdot (2\hat{\imath} - \hat{\jmath} + \hat{k}) = 3$ and passing through the origin.

No answer yet.

2018 · 4 marks · DerivationOpen: If are three non-collinear points with position vectors , respectively, then…
If $A, B, C$ are three non-collinear points with position vectors $\vec{a}, \vec{b}, \vec{c}$, respectively, then show that the length of the perpendicular from $C$ on $AB$ is $\frac{|(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})|}{|\vec{b} - \vec{a}|}$.

Given: $A, B, C$ are three non-collinear points with position vectors $\vec{a}, \vec{b}, \vec{c}$, respectively

To show: Length of the perpendicular from $C$ on $AB$ is $\frac{|(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})|}{|\vec{b} - \vec{a}|}$

No answer yet.

Questions on other pages on Vectors

Other Mathematics chapters