Download App

Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors5 Marks
Question
Prove that the given vectors are non-coplanar:
$3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\ 2\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}$ and $7\hat{\text{i}}-\hat{\text{j}}+23\hat{\text{k}}$
✓

Answer

We know that, Three vectors are coplanar if one of them vector can be expressed as the linear combination of the other two. Let, $\big(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)\\=\text{x}\big(2\hat{\text{i}}-\hat{\text{j}}+7\hat{\text{k}}\big)+\text{y}\big(7\hat{\text{i}}-\hat{\text{j}}+23\hat{\text{k}}\big)$ $=2\text{x}\hat{\text{i}}-\text{x}\hat{\text{j}}+7\text{x}\hat{\text{k}}+7\text{y}\hat{\text{i}}-\text{y}\hat{\text{j}}+23\text{y}\hat{\text{k}}$ $\big(3\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)\\=\big(2\text{x}+7\text{y}\big)\hat{\text{i}}+\big(-\text{x}-\text{y}\big)\hat{\text{j}}+\big(7\text{x}+23\text{y}\big)\hat{\text{k}}$ Equating the coefficient of LHS and RHS, 2x + 7y = 3 .....(i) -x - y = 1 .....(ii) 7x + 23y = -1 .....(iii) For solving (i) and (ii), Add (i) and 2 × (ii),
$\text{y}=\frac{5}5$ $\text{y}=1$ Put the value of y in equation (i), $2\text{x}+7\text{y}=3$ $2\text{x}+7(1)=3$ $2\text{x}+7=3$ $2\text{x}=3-7$ $2\text{x}=-4$ $\text{x}=\frac{-4}2$ $\text{x}=-2$ Put the value of x and y in equation (iii), $7\text{x}+23\text{y}=-1$ $7(2)+23(1)=-1$ $14+23=-1$ $37=-1$ $\text{LHS}\neq\text{RHS}$ The value of x and y do not satisfy the equation (iii), Hence, vectors are non-coplanar.

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 "5 Marks Questions" from Algebra of Vectors→Algebra of Vectors — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Find the angle of intersecting of the following curves:
$\text{y}=4-\text{x}^2\text{ and }\text{y}=\text{x}^2$
→
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1\text{ at }(\text{x}_1,\text{y}_1)$
→
If f'(x) = x + b, f'(1) = 5, f'(2) = 13, find f'(x).
→
Evaluate the following integrals:
$\int_{0}^\limits{1}\text{x}\log(1+2\text{x})\text{dx}$
→
Find the area included between the parabolas $y^2 = 4ax$ and $x^2 = 4 by$.
→
Differentiate the following functions with respect to x:
$\log\sqrt{\frac{\text{x}-1}{\text{x}+1}}$
→
If $\text A = \begin{bmatrix}1&2&-3\\5&0&2\\1&-1&1\end{bmatrix},\ \text B = \begin{bmatrix}3&-1&2\\4&2&5\\2&0&3\end{bmatrix} \text{and}\ \text C = \begin{bmatrix}4&1&2\\0&3&2\\1&-2&3\end{bmatrix},$ then compute (A + B) and (B - C).
Also, verify that A + (B - C) = (A + B) - C.
→
If A and B are two events such that,
$\text{P(A)}=\frac{1}{3},\text{P(B)}=\frac{1}{4}$ and $\text{P}(\text{A}\cap\text{B})=\frac{5}{12},$ then find P(A|B) and P(B|A).
→
Show that the function defined by f(x) = | cos x | is a continuous function.
→
Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?
→